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MAT 2125Elementary Real Analysis
Chapter 5Differential and Integral Calculus
P. Boily (uOttawa)
Winter 2022
P. Boily (uOttawa)
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Overview
We have spent a fair amount of time and energy on concepts like the limit,continuity, and uniform continuity, with the goal of making differential andintegral calculus sound.
In this chapter, we
introduce the concepts of differentiability and Riemann-integrability forfunctions, and
prove a number of useful calculus results.
P. Boily (uOttawa) 1
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Outline
5.1 â Differentiation (p.3)ab Mean Value Theorem (p.20)
ab Taylor Theorem (p.28)
ab Relative Extrema (p.35)
5.2 â Riemann Integral (p.43)ab Riemann Criterion (p.59)
ab Properties of the Riemann Integral (p.68)
ab Fundamental Theorem of Calculus (p.88)
ab Evaluation of Integrals (p.96)
5.3 â Exercises (p.101)
P. Boily (uOttawa) 2
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5.1 â Differentiation
Let I â R be an interval, f : I â R, and c â I. The real number L is thederivative of f at c, denoted by f âČ(c) = L, if
âΔ > 0,âΎΔ > 0 s.t. x â I and 0 < |xâc| < ΎΔ =ââŁâŁâŁâŁf(x)â f(c)xâ c
â LâŁâŁâŁâŁ < Δ.
In that case, we say that f is differentiable at c. This definition simplystates that f âČ(c) exists if lim
xâcf(x)âf(c)
xâc exists, and that, in that case,
f âČ(c) = limxâc
f(x)â f(c)xâ c
.
P. Boily (uOttawa) 3
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
While f âČ(c) â R (if it exists), f âČ : I â R is the derivative function.
Example: Let f : I â R be defined by f(x) = x3. Set c â I. Then
f âČ(c) = limxâc
f(x)â f(c)xâ c
= limxâc
x3 â c3
xâ c= limxâc
(x2 + cx+ c2) = 3c2.
The corresponding derivative function is f âČ : I â R, f âČ(x) = 3x2.
Theorem 39. If f : I â R has a derivative at c, then f is continuous at c.
Proof. Let x, c â I, x 6= c. Then f(x)â f(c) =(f(x)âf(c)
xâc
)(xâ c) and so
limxâc
(f(x)â f(c)) = limxâc
f(x)â f(c)xâ c
· limxâc
(xâ c),
P. Boily (uOttawa) 4
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
if all the limits exist. But xâ câ 0 when xâ c, and
limxâc
f(x)â f(c)xâ c
= f âČ(c)
by hypothesis, so
limxâc
(f(x)â f(c)) = f âČ(c) = 0 = 0 =â limxâc
f(x) = f(c),
which means that f is continuous at c. ïżœ
However, the converse of Theorem 38 does not always hold. For instance,the function | · | : R â R is continuous at x = 0, but it has no derivativethere as |x|/x has no limit when xâ 0.
P. Boily (uOttawa) 5
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Continuity is a necessary condition for differentiability, but it is not sufficient.
In fact, it is not too difficult to find functions that are continuouseverywhere, but nowhere differentiable.
Example: Weierstrass provided the first example of such a function in1872: f : Râ R defined by
f(x) =ânâN
cos(3nx)
2n.
That it took so long to find an example is mostly due to the fact that thedefinition of a function has evolved a fair amount over the last 200 years.
P. Boily (uOttawa) 6
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Theorem 40. Let I be an interval, c â I, α â R, and f, g : I â R bedifferentiable at c, with g(c) 6= 0. Then
(a) αf is differentiable at c and (αf)âČ(c) = αf âČ(c);
(b) f + g is differentiable at c and (f + g)âČ(c) = f âČ(c) + gâČ(c);
(c) fg is differentiable at c and (fg)âČ(c) = f âČ(c)g(c) + f(c)gâČ(c);
(d) f/g is differentiable at c and (f/g)âČ(c) =f âČ(c)g(c)â f(c)gâČ(c)
[g(c)]2.
Proof. In all instances, we compute the limit of the differential quotient,taking into account the fact that f and g are differentiable at c.
P. Boily (uOttawa) 7
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
(a) If αf is differentiable at c, then
(αf)âČ(c) = limxâc
(αf)(x)â (αf)(c)
xâ c
= limxâc
α(f(x)â f(c))xâ c
= α limxâc
f(x)â f(c)xâ c
.
But f is differentiable at c, so the last limit exists, validating the stringof equalities, and is equal to f âČ(c).
Hence (αf)âČ(c) = αf âČ(c).
P. Boily (uOttawa) 8
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
(b) If f + g is differentiable at c, then
(f + g)âČ(c) = limxâc
(f + g)(x)â (f + g)(c)
xâ c
= limxâc
f(x) + g(x)â f(c)â g(c)xâ c
= limxâc
f(x)â f(c)xâ c
+ limxâc
g(x)â g(c)xâ c
.
But both f, g is differentiable at c, so the sum of limits exists, validatingthe string of equalities, and is equal to f âČ(c) + gâČ(c).
Hence (f + g)âČ(c) = f âČ(c) + gâČ(c).
P. Boily (uOttawa) 9
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
(c) If fg is differentiable at c, then
(fg)âČ(c) = limxâc
(fg)(x)â (fg)(c)
xâ c= limxâc
f(x)g(x)â f(c)g(c)xâ c
= limxâc
f(x)g(x)â f(c)g(x) + f(c)g(x)â f(c)g(c)xâ c
= limxâc
f(x)â f(c)xâ c
g(x) + limxâc
f(c)g(x)â g(c)xâ c
= limxâc
f(x)â f(c)xâ c
· limxâc
g(x) + f(c) limxâc
g(x)â g(c)xâ c
But both f, g is differentiable at c, so the differential quotient limits exist,validating the string of equalities. Furthermore, g is continuous at c,being differentiable at c (acccording to Theorem 39).
P. Boily (uOttawa) 10
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Hence
(fg)âČ(c) = f âČ(c) limxâc
g(x) + f(c)gâČ(c) = f âČ(c)g(c) + f(c)gâČ(c).
(d) Since g is continuous at c (being differentiable at c) and g(c) 6= 0, â aninterval J â I such that c â J an g 6= 0 on J . If f/g is differentiable atc, then
(f/g)âČ(c) = limxâc
(f/g)(x)â (f/g)(c)
xâ c= limxâc
f(x)/g(x)â f(c)/g(c)xâ c
= limxâc
f(x)g(c)â f(c)g(x)g(x)g(c)(xâ c)
= limxâc
f(x)g(c)â f(c)g(c) + f(c)g(c)â f(c)g(c)g(x)g(c)(xâ c)
,
P. Boily (uOttawa) 11
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
so that
(f/g)âČ(c) = limxâc
1
g(x)g(c)
[f(x)â f(c)
xâ cg(c)â f(c)g(x)â g(c)
xâ c
]= limxâc
1
g(x)g(c)·[limxâc
f(x)â f(c)xâ c
g(c)â f(c) limxâc
g(x)â g(c)xâ c
].
But both f, g is differentiable at c, so the differential quotient limitsexist, validating the string of equalities.
Furthermore, g is continuous at c, being differentiable at c (cf. Theorem 39),and g 6= 0 on J , so that 1
g(x) â1g(c) when xâ c. Thus
(f/g)âČ(c) =f âČ(c)g(c)â f(c)gâČ(c)
[g(c)]2. ïżœ
P. Boily (uOttawa) 12
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
By mathematical induction, we can easily show that
[ nâi=1
fi
]âČ(c) =
nâi=1
f âČi(c) and[ nâi=1
fi
]âČ(c) =
nâi=1
(âj 6=i
fj(c))f âČi(c),
if f1, . . . , fn are all differentiable at c. In particular, if f1 = · · · = fn, then
(fn)âČ(c) = nfnâ1(c) · f âČ(c).
Consider the identity function f . Then for c â R,
f âČ(c) = limxâc
xâ cxâ c
= 1, =â (fn)âČ(x) = nfnâ1(x) · f âČ(x) = nxnâ1
for all x â R, n â N, which can be extended to n â Z using Theorem 40d.
P. Boily (uOttawa) 13
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Theorem 41. (Caratheodory)Let I = [a, b] and f : I â R. Then f is differentiable at c â I if and onlyif âÏc : I â R, continuous at c such that f(x)â f(c) = Ïc(x)(xâ c), forall x â I. In that case, Ïc(c) = f âČ(c).
Proof. Let c â I. Assume that f âČ(c) exists. Define Ïc : I â R by
Ïc(x) =
{f(x)âf(c)
xâc , x 6= c
f âČ(c), x = c
Then Ïc is continuous at c since
limxâc
Ïc(x) = limxâc
f(x)â f(c)xâ c
= f âČ(c) = Ïc(c).
P. Boily (uOttawa) 14
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
If x = c, then f(x) = f(c) and
f(x)â f(c) = f(c)â f(c) = 0 = Ïc(c)(câ c) = Ïc(x)(xâ c).
If x 6= c and x â I, then, by definition,
f(x)â f(c) = Ïc(x)(xâ c).
Assume now that âÏc : I â R, continuous at c, and such that
(x)â f(c) = Ïc(x)(xâ c), for all x â I.
If x 6= c, then
Ïc(x) =f(x)â f(c)
xâ c
P. Boily (uOttawa) 15
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
and, since Ïc is continuous at c,
Ïc(c) = limxâc
Ïc(x) = limxâc
f(x)â f(c)xâ c
exists. Then Ïc(c) = f âČ(c) and f is differentiable at c. ïżœ
It is important to recognize that Ïc is not, as a function, the same as f âČ,in general â it is only at c that they can be guaranteed to coincide,although in certain cases (such as when f is a linear function), f âČ(x) = Ïc(x)for all c in I.
Caretheodoryâs Theorem can be used to prove the Chain Rule of calculus.
P. Boily (uOttawa) 16
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Theorem 42. (Chain Rule)Let I, J be closed bounded intervals, g : I â R and f : I â R be functionssuch that f(J) â I, and let c â J , with d = f(c). If f is differentiableat c and g is differentiable at d, then the composition g ⊠f : J â R isdifferentiable at c and (g ⊠f)âČ(c) = gâČ(f(c))f âČ(c) = gâČ(d)f âČ(c).
Proof. Since f âČ(c) exists, Caratheodoryâs Theorem implies that Ïc : J â Rsuch that Ïc is continuous at c â J with
Ïc(c) = f âČ(c), and f(x)â f(c) = Ïc(x)(xâ c), for all x â J.
Since gâČ(d) exists, âÏd : I â R such that Ïd is continuous at d â I, with
Ïd(d) = gâČ(d), and g(y)â g(d) = Ïd(y)(y â d), for all y â I.
P. Boily (uOttawa) 17
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Thus, if y = f(x) and d = f(c), we have
(g ⊠f)(x)â (g ⊠f)(c) = g(f(x))â g(f(c)) = Ïd(f(x))(f(x)â f(c))
= Ïd(f(x))Ïc(x)(xâ c) =[(Ïd ⊠f)(x) · Ïc(x)
](xâ c),
for all x â J such that f(x) â I.
But (Ïd ⊠f) · Ïc is continuous at c, being the product of two functionswhich are continuous at c.
According to Caratheodoryâs Theorem, (Ïd ⊠f)(c) ·Ïc(c) = (g ⊠f)âČ(c). But
(Ïd ⊠f)(c) · Ïc(c) = Ïd(f(c))Ïc(c) = gâČ(f(c))f âČ(c) = gâČ(d)f âČ(c),
which completes the proof. ïżœ
P. Boily (uOttawa) 18
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Example: Suppose that f : I â R is differentiable at c and that f, f âČ 6= 0on I. If h is defined by h(y) = 1
y, y 6= 0, then hâČ(y) = â 1y2
. Thus
(1/f)âČ(x) = (h ⊠f)âČ(x) = hâČ(f(x)) · f âČ(x) = â f âČ(x)
(f(x))2, for all x â I.
Example: Let g = | â |. Then gâČ(c) = sgn(c) for all c 6= 0. Indeed,
limxâc
|x| â |c|xâ c
= limxâc
{xâcxâc, c > 0
âxâcxâc, c < 0=
{1, c > 0
â1, c < 0= sgn(c),
but gâČ(0) does not exist (even though sgn(0) = 0). If f : [a, b] â R isdifferentiable, the Chain Rule states that |f |âČ(x) = sgn(f(x)) · f âČ(x). Whathappens if f(c) = 0? Is |f | differentiable at c?
P. Boily (uOttawa) 19
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5.1.1 â Mean Value Theorem
Let I be an interval. A function f : I â R has a relative maximum atc â I if âÎŽ > 0 such that
f(x) †f(c), âx â VÎŽ(c) = (câ ÎŽ, c+ ÎŽ);
it has a relative minimum at c â I if âÎŽ > 0 such that
f(x) â„ f(c), âx â VÎŽ(c) = (câ ÎŽ, c+ ÎŽ).
If f has either a relative maximum or a relative minimum at c, we say thatit has a relative extremum at c.
P. Boily (uOttawa) 20
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Note that the definition of relative extremum makes no mention of continuityor differentiability.
Theorem 43. Let f : [a, b]â R, c â (a, b). If f has a relative extremumat c and if f is differentiable at c, then f âČ(c) = 0.
Proof. Without loss of generality, assume that f has a relative maximumat c; the proof for a relative minimum follows the same lines.
Let ÎŽ be the quantity whose existence is guaranteed by the definition:
f(x) †f(c), âx â VÎŽ.
If f âČ(c) > 0, then âÎŽ > 0 such that f(x)âf(c)xâc > 0 whenever 0 < |xâ c| < ÎŽ.
P. Boily (uOttawa) 21
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Indeed, according to the definition of the derivative, if Δ = 12fâČ(c) > 0,
âΎΔ > 0 such that âŁâŁâŁâŁf(x)â f(c)xâ câ f âČ(c)
âŁâŁâŁâŁ < Δ =1
2f âČ(c)
whenever 0 < |xâ c| < ΎΔ. Set ÎŽ = min{ΎΔ, ÎŽ}. Then
â12f âČ(c) <
f(x)â f(c)xâ c
â f âČ(c) < 1
2f âČ(c), whenever 0 < |xâ c| < ÎŽ,
and so
0 <1
2f âČ(c) <
f(x)â f(c)xâ c
, whenever 0 < |xâ c| < ÎŽ.
P. Boily (uOttawa) 22
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
But if x â VÎŽ(c) with x > c, then
f(x)â f(c) = (xâ c)ïžž ïž·ïž· ïžž>0
· f(x)â f(c)xâ cïžž ïž·ïž· ïžž>0
> 0,
and so f(x) > f(c), which contradicts the fact that f has a relativemaximum at c. Thus, f âČ(c) 6> 0.
We can prove that f âČ(c) 6< 0 using a similar argument. As neitherf âČ(c) > 0 nor f âČ(c) < 0, we must have f âČ(c) = 0. ïżœ
This result justifies the common practice o looking for relative extremaat roots of the derivative. Note that c is not an endpoint of the interval,and so we must also included them in the search for extrema. What happensif f is not differentiable at c in Theorem 43?
P. Boily (uOttawa) 23
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
The next theorem has far-reaching consequences.
Theorem 44. (Rolle)Let f : [a, b] â R be continuous on [a, b] and differentiable on (a, b). Iff(a) = f(b) = 0, âc â (a, b) such that f âČ(c) = 0.
Proof. If f ⥠0 on [a, b], then the conclusion holds for any c â (a, b).
If âxâ such that f(xâ) 6= 0, we may suppose, without loss of generality, thatf(xâ) > 0. According to the Max/Min Theorem, f reaches its maximum
sup{f(x) | x â [a, b]} > 0
somewhere in [a, b]. But since f(a) = f(b) = 0, the maximum must bereached in (a, b). Denote that point by c. Then f âČ(c) exists and since fhas a realative maximum at c, Theorem 43 implies that f âČ(c) = 0. ïżœ
P. Boily (uOttawa) 24
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
The Mean Value Theorem is an easy corollary of Rolleâs Theorem.
Theorem 45. (Mean Value Theorem)Let f : [a, b] â R be continuous on [a, b]. If f is differentiable on (a, b),âc â (a, b) such that f(b)â f(a) = f âČ(c)(bâ a).
Proof. Let Ï : [a, b]â R be defined by
Ï(x) = f(x)â (a)â f(b)â f(a)bâ a
(xâ a).
Then
Ï(a) = f(a)â f(a)â f(b)â f(a)bâ a
(aâ a) = 0
Ï(b) = f(b)â f(a)â f(b)â f(a)bâ a
(bâ a) = f(b)â f(a)â (f(b)â f(a)) = 0.
P. Boily (uOttawa) 25
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
But Ï is continuous on [a, b] as f and x 7â xâ a are continuous on [a, b].According to Rolleâs Theorem, âc â (a, b) such that ÏâČ(c) = 0. But
ÏâČ(x) = f âČ(x)â f(b)â f(a)bâ a
,
so that f âČ(c)â f(b)âf(a)bâc = 0, which completes the proof. ïżœ
Theorem 46. Let f : [a, b]â R be continuous on [a, b] and differentiableon (a, b). If f âČ âĄ 0 on (a, b), then f is constant on [a, b].
Proof. Let x â (a, b]. According to the MVT, âc â (a, x) such that
f(x)â f(a) = f âČ(c)(xâ a).
But f âČ(c) = 0, so that f(x)â f(a) = 0 for all x â [a, b]. ïżœ
P. Boily (uOttawa) 26
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
An illustration of Rolleâs Theorem (left); Mean Value Theorem (right).
P. Boily (uOttawa) 27
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5.1.2 â Taylor Theorem
Taylorâs Theorem is used extensively in applications. It is, in a way, anextension of the Mean Value Theorem to higher order derivatives.
We can naturally obtain the higher-order derivatives of a function fby formally applying the differentiation rules repeatedly.
Hence, f (2) = f âČâČ = (f âČ)âČ, f (3) = f âČâČâČ = (f âČâČ)âČ = ((f âČ)âČ)âČ, etc.
Suppose f = f (0) can be differentiated n times at x = x0. The nthTaylor polynomial of f at x = x0 is
Pn(x; f, x0) =
nâi=0
f (i)(x0)
i!(xâ x0)i.
P. Boily (uOttawa) 28
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Theorem 47. (Taylor)Let n â N and f : [a.b] â R be such that f and its derivativesf âČ, f âČâČ, . . . , f (n) are continuous on [a, b], and f (n+1) exists on (a, b). Ifx0 â [a, b], then for all x 6= x0 â [a, b], âc between x and x0 such that
f(x) = Pn(x; f, x0) +f (n+1)(c)
(n+ 1)!(xâ x0)n+1.
Proof. Let x â [a, b]. If x0 < x, set J = [x0, x]. Otherwise, set J = [x, x0].
Let F : J â R be defined by
F (t) = f(x)âPn(t; f, x) = f(x)âf(t)âf âČ(t)(xât)â· · ·âf(n)(t)
n!(xât)n.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Note that F is continuous on J as f and its n higher-order derivatives arecontinuous on J , and that
F âČ(t) = âf âČ(t)â[f âČâČ(t)(xâ t)â f âČ(t)
]â[f âČâČâČ(t)
2!(xâ t)2 â f âČâČ(t)(xâ t)
]â · · ·â
â[f (n+1)(t)
n!(xâ t)n â f (n)(t)(xâ t)nâ1
].
Thus F âČ(t) = âf(n+1)(t)n! (xâ t)n. Let G : J â R be defined by
G(t) = F (t)â(xâ txâ x0
)n+1
F (x0).
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Then
G(x0) = F (x0)â(xâ x0xâ x0
)n+1
F (x0) = 0
G(x) = F (x)â(xâ xxâ x0
)n+1
F (x0) = F (x).
But
F (x) = f(x)â f(x)â f âČ(x)(xâ x)â · · · â f(n)(x)
n!(xâ x)n = 0.
Thus G(x) = 0. Note that G is continuous on J .
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Furthermore, G is differentiable on J since
GâČ(t) = F âČ(t) +(n+ 1)
xâ x0
(xâ txâ x0
)nF (x0)
= âf(n+1)(t)
n!(xâ t)n + (n+ 1)
xâ x0
(xâ txâ x0
)nF (x0).
As G satisfies the hypotheses of Rolleâs Theorem, âc between x and x0such that GâČ(c) = 0. Thus
f (n+1)(c)
n!(xâ c)n = (n+ 1)
(xâ c)n
(xâ x0)n+1F (x0) =â
F (x0) =f (n+1)(c)
n!(n+ 1)
(xâ c)n
(xâ c)n(xâ x0)n+1 =
f (n+1)(c)
(n+ 1)!(xâ x0)n+1.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
But
F (x0) = f(x)â Pn(x0; f, x) =â f(x) = Pn(f ;x0) + F (x0),
which completes the proof. ïżœ
Example: Use Taylorâs Theorem with n = 2 to approximate 4â1 + x
near x0 = 0 (for x > â1).
Solution. Let f(x) = (1 + x)1/4. Then
f âČ(x) =1
4(1+x)â3/4, f âČâČ(x) = â 3
16(1+x)â7/4, f âČâČâČ(x) =
21
64(1+x)â11/4
are all continuous in closed intervals [âa, a], 1 > a > 0, so Taylorâs Theoremcan be brought to bear on the situation.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Note that f(0) = 1, f âČ(0) = 14 and f âČâČ(0) = â 3
16. According to TaylorâsTheorem, for every x â [âa, a], 1 > a > 0, âc between x and 0 such that
f(x) = P2(x; f, 0) +f âČâČâČ(c)
3!x3 = 1 +
1
4xâ 3
32x2 +
7
128(1 + c)11/4x3.
For instance, 4â1.4 can be approximated by
f(0.4) â P2(0.4) = 1 + 14(0.4)â
332(0.4)
2 â 1.085.
Moreover, since c â (0, 0.4),
f âČâČâČ(c)6 (0.4)3 = 7
128(1 + c)â11/4(0.4)3 †7128(0.4)
3 = 0.0035,
so | 4â1.4 â 1.085| †0.0035, which is to say that the approximation is
correct to 2 decimal places.
P. Boily (uOttawa) 34
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5.1.3 â Relative Extrema
We end this section by giving a characterization of relative extrema usingthe derivative.
A function f : I â R is increasing (resp. decreasing) if
f(x1) †f(x2), (resp. f(x1) â„ f(x2)) âx1 †x2 â I.
If the inequalities are strict, then the function is strictly increasing (resp.strictly decreasing).
A function that is either increasing or decreasing (exclusively) is monotone.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
If the function is also differentiable, then a link exists.
Theorem 48. Let f : [a, b]â R be continuous on [a, b], differentiable on(a, b). Then f is increasing on [a, b] if and only if f âČ â„ 0 on (a, b).
Proof. Suppose f is increasing and let c â (a, b). For all x < c in (a, b),we have f(x) †f(c); for all x > c in (a, b), we have f(x) â„ f(c). Thus
f(x)â f(c)xâ c
â„ 0, for all x 6= c â (a, b).
Since f is differentiable at c, we must have
f âČ(c) = limxâc
f(x)â f(c)xâ c
â„ 0.
As c is arbitrary, we have f âČ(x) â„ 0 for all x â (a, b).
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
If, conversely, f âČ(x) â„ 0 for all x â (a, b), let x1 < x2 â [a, b]. By the MeanValue Theorem, âc â (x1, x2) such that
f(x2)â f(x1) = f âČ(c)(x2 â x1).
Since f âČ(c) â„ 0 an x2 > x1, then
f(x2)â f(x1)x2 â x1
= f âČ(c) â„ 0 =â f(x2)â f(x1) â„ 0 =â f(x2) â„ f(x1),
which is to say, f is increasing on [a, b]. ïżœ
Theorem 48 holds for decreasing functions as well (after having madethe obvious changes to the statement). If we switch to strictly monotonefunctions, only one direction holds in all cases â which one?
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
The next result is a celebrated result from calculus.
Theorem 49. (First Derivative Test)Let f be continuous on [a, b] and let c â (a, b). Suppose f is differentiableon (a, c) and on (b, c), but not necessarily at c. Then
(a) if âVÎŽ(c) â [a, b] such that f âČ(x) â„ 0 for c â ÎŽ < x < c and f âČ(x) †0for c < x < c+ ÎŽ, then f has a relative maximum at c;
(b) if âVÎŽ(c) â [a, b] such that f âČ(x) †0 for c â ÎŽ < x < c and f âČ(x) â„ 0for c < x < c+ ÎŽ, then f has a relative minimum at c.
Proof. We only prove part (a); the proof for (b) follows the same lines.
P. Boily (uOttawa) 38
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
If x â (câÎŽ, c), the Mean Value Theorem states that âcx â (x, c) such that
f(c)â f(x) = f âČ(cx)ïžž ïž·ïž· ïžžâ„0
(câ x)ïžž ïž·ïž· ïžžâ„0
â„ 0,
so that f(x) †f(c) for all x â (câ ÎŽ, c).
If x â (c, c + ÎŽ), the Mean Value Theorem states that âcx â (c, x)such that
f(c)â f(x) = f âČ(cx)ïžž ïž·ïž· ïžžâ€0
(câ x)ïžž ïž·ïž· ïžžâ€0
â„ 0,
so that f(x) †f(c) for all x â (c, c+ ÎŽ).
Combining these two statements with the fact that f(c) †f(c), weobtain f(x) †f(c) for all x â VÎŽ(c), so f has a relative maximum at c. ïżœ
P. Boily (uOttawa) 39
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
The converse of the First Derivative Test is not necessarily true. Forinstance, the function defined by
f(x) =
{2x4 + x4 sin(1/x), x 6= 0
0 x = 0
has an absolute minimum at x = 0, but it has derivatives of either sign oneither side of any neighbourhood of x = 0.
We end this section with a rather surprising result.
Theorem 50. (Darboux)Let f : [a, b]â R be differentiable, continuous on [a, b] and let k be strictlyconfined between f âČ(a) and f âČ(b). Then âc â (a, b) with f âČ(c) = k.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Proof. Without loss of generality, assume f âČ(a) < k < f âČ(b).
Define g : [a, b] â R by g(x) = kx â f(x); g is then continuous anddifferentiable on [a, b] given that both f and x 7â kx also are.
According to the Max/Min Theorem, g reaches its maximum value atsome c â [a, b]. However, gâČ(a) = k â f âČ(a) > 0, so that c 6= a, andgâČ(b) = k â f âČ(b) < 0, so that c 6= b.
Hence gâČ(c) = 0 for some c â (a, b), according to Theorem 43, and sof âČ(c) = k. ïżœ
Darbouxâs Theorem states that the derivative of a continuous function,which needs not be continuous, nevertheless satisfies the intermediate valueproperty.
P. Boily (uOttawa) 41
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
There are a number of other results which could be shown aboutdifferentiable functions, but that are left as exercises. Let f : (a, b)â R bedifferentiable on (a, b), with f âČ(x) 6= 0. Then
f is monotone on (a, b) and f((a, b)) is an open interval (α, ÎČ);
f has an inverse fâ1 : (α, ÎČ)â R such that
fâ1(f(x)) = x, f(fâ1(y)) = y, âx â (a, b), y â (α, ÎČ),
fâ1 is differentiable on (α, ÎČ), with
(fâ1)âČ(y) =1
f âČ(fâ1(y)), ây â (α, ÎČ).
P. Boily (uOttawa) 42
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5.2 â Riemann Integral
Calculus as a discipline only took flight after Newton announced his theoryof fluxions. With Leibnizâ independent discovery that the reversal of theprocess for fining tangents lea to areas under curves, integration was born.
Riemann was the first to study integration as a process separate fromdifferentiation.
We start by studying the integration of a functions R â R. Later on,you will tackle integration of functions Rn â R; and, eventually, offunctions Rn â Rn.
P. Boily (uOttawa) 43
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Let I = [a, b]. A partition P of [a, b] is a subset P = {x0, . . . , xn} â Isuch that
a = x0 < x1 < · · · < xnâ1 < xn = b.
If f : I â R is bounded and P is a partition of I, the sum
L(P ; f) =
nâi=1
mi(xi â xiâ1) <â, U(P ; f) =
nâi=1
Mi(xi â xiâ1) <â,
where
mi = inf{f(x) | x â [xiâ1, xi]}, Mi = sup{f(x) | x â [xiâ1, xi]}, 1 †i †n
are the lower and the upper sum of f corresponding to P , respectively.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
If f : I â R+0 , we can give a graphical representation of these sums.
L(P ; f) is the area of the union of the rectangles with base [xkâ1, xk] andheight mk, and U(P ; f) is the area of the union of the rectangles with base[xkâ1, xk] and height Mk.
P. Boily (uOttawa) 45
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
In general, a partition Q of I is a refinement of a partition P of I if P â Q.
Example: Both P = {0, 1, 4, 10} and Q = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10} arepartitions of [0, 10]; since Q â P , Q is a refinement of P .
Lemma 1. Let I = [a, b] and f : I â R be bounded. Then
(a) L(P ; f) †U(P ; f) for any partition P of I;
(a) L(P ; f) †L(Q; f) and U(Q; f) †L(Q; f) for any refinement Q â Pof I, and
(a) L(P1; f) †U(P2; f) for any pair of partitions P1, P2 of I.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Proof.
(a) Let P = {x0, . . . , xn} be a partition of I. Since
mi = inf{f(x) | x â [xiâ1, xi]} †sup{f(x) | x â [xiâ1, xi]} =Mi
for all 1 †i †n, then
L(P ; f) =
nâi=1
mi(xi â xiâ1ïžž ïž·ïž· ïžž>0
) â€nâi=1
Mi(xi â xiâ1ïžž ïž·ïž· ïžž>0
) = U(P ; f).
(b) Let Q = {y0, . . . , ym} be a refinement of P = {x0, . . . , xn}. SetIi = [xiâ1, xi] and Ij = [yjâ1, yj], for 1 †i †n and 1 †j †m.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Write mi = inf{f(x) | x â Ii} and mj = inf{f(x) | x â Ij} and fix1 †1 †n. Then âj, k such that
Ii = Ij+1 âȘ · · · âȘ Ij+k =kâ`=1
Ij+`.
Then
mi(xi â xi â 1) = mi(yj + k â yj) = mi(yj+1 â yj + · · ·+ yj+k â yj+kâ1)= mi(yj+1 â yj) + · · ·+mi(yj+k â yj+kâ1)
=
kâ`=1
mi(yj+` â yj+`â1) â€kâ`=1
mj+`(yj+` â yj+`â1)
since Ij+` â Ii for all ` = 1, . . . , k.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Hence
L(P ; f) =
nâi=1
mi(xi â xiâ1) â€mâj=1
mj(yj â yjâ1) = L(Q; f).
The proof for U(P ; f) â„ U(Q; f) follows a similar path.
(c) Let P1, P2 be partitions of I. Set Q = P1 âȘ P2. Then Q is a refinementof both P1 and P2. By the results prove in parts (a) and (b) of thislemma, we have
L(P1; f) †L(Q; f) †U(Q; f) †U(P2; f),
which completes the proof. ïżœ
P. Boily (uOttawa) 49
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Let I = [a, b] and f : I â R be bounded. The lower integral of f on I isthe number
L(f) = sup{L(P ; f) | P a partition of I}.The upper integral of f on I is the number
U(f) = inf{U(P ; f) | P a partition of I}.
Since f is bounded on I, âm,M such that m †f(x) †M for all x â I.Consider the trivial partition P0 = {a, b}. Since any partition P of I is arefinement of P0, we thus have
L(P ; f) †U(P0; f) â€M(bâ a) and U(P ; f) â„ L(P0; f) â„ m(bâ a).
Thus L(f), U(f) exist, by completeness. But we can say more.
Theorem 51. Let f : [a, b]â R be bounded. Then L(f) †U(f).
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Proof. Let P1, P2 be partitions of [a, b]. Then L(P1; f) †U(P2; f). If wefix P2, U(P2; f) is an upper bound for
A = {L(P1; f) | P1 is a partition of [a, b]}.
Since L(f) = sup(A) and since P2 was arbitrary, L(f) is a lower bound for
B = {U(P2; f) | P2 is a partition of [a, b]}.
Thus L(f) †inf(B) = U(f). ïżœ
When L(f) = U(f), we say that f is Riemann-integrable on [a, b];the integral of f on [a, b] is the real number
L(f) = U(f) =
â« b
a
f =
â« b
a
f(x) dx.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
By convention, we defineâ« baf = â
â« abf when b < a. Note that
â« aaf = 0
for all bounded functions f .
Example: Show directly that the function defined by h(x) = x2 is Riemann-
integrable on [a, b], b > a â„ 0. Furthermore show thatâ« bah = b3âa3
3 .
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Proof. Let Pn ={xi = a+ bâa
n · i | i = 0, . . . , n}
be a partition of [a, b].Set mi = inf{h(x) | x â [xiâ1, xi]}, for i = 1, . . . , n.
With this notation, we have
L(Pn;h) =
nâi=1
mi(xi â xiâ1) =bâ an
nâi=1
mi.
But hâČ(x) = 2x â„ 0 when x â„ 0, and so h is increasing on [a, b].
Consequently, for i = 1, . . . , n, we have
mi = x2iâ1 =
(a+
bâ an
(iâ 1)
)2
= a2+2a(bâ a)
n(iâ1)+(bâ a)2
n2(iâ1)2.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
The lower sum of h associated to Pn is thus
L(Pn;h) =bâ an
nâi=1
(a2 + 2
a(bâ a)n
(iâ 1) +(bâ a)2
n2(iâ 1)2
)
=na2(bâ a)
n+
2a(bâ a)2
n2
nâi=1
(iâ 1) +(bâ a)3
n3
nâi=1
(iâ 1)2
= a2(bâ a) + 2a(bâ a)2
n2· n(nâ 1)
2+
(bâ a)3
n3· n(nâ 1)(2nâ 1)
6
= a2(bâ a) + a(bâ a)2(1â 1
n
)+
(bâ a)3
6
(1â 1
n
)(2â 1
n
).
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
But for the lower sum of h on [a, b], we have
L(h) = sup{L(P ;h) | P â P([a, b])} â„ supnâN{L(Pn;h)}
= supnâN
{a2(bâ a) + a(bâ a)2
(1â 1
n
)+ (bâa)3
6
(1â 1
n
) (2â 1
n
)}= limnââ
[a2(bâ a) + a(bâ a)2
(1â 1
n
)+ (bâa)3
6
(1â 1
n
) (2â 1
n
)]= a2(bâ a) + a(bâ a)2 + (bâ a)3
6· 2 =
b3 â a3
3
Similarly, we can show that
U(Pn;h) = a2(bâ a) + a(bâ a)2(1 +
1
n
)+
(bâ a)3
6
(1 +
1
n
)(2 +
1
n
).
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
But for the upper sum of h on [a, b], we have
U(h) = inf{U(P ;h) | P â P([a, b])} †infnâN{U(Pn;h)}
= infnâN
{a2(bâ a) + a(bâ a)2
(1 + 1
n
)+ (bâa)3
6
(1 + 1
n
) (2 + 1
n
)}= limnââ
[a2(bâ a) + a(bâ a)2
(1 + 1
n
)+ (bâa)3
6
(1 + 1
n
) (2 + 1
n
)]= a2(bâ a) + a(bâ a)2 + (bâ a)3
6· 2 =
b3 â a3
3
Thus b3âa33 †L(h) †U(h) †b3âa3
3 and so L(h) = U(h) =â« bah = b3âa3
3 ,which completes the proof. ïżœ
P. Boily (uOttawa) 56
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Example: Show directly that the Dirchlet function defined by
f(x) =
{1, x â Q0, x 6â Q
is not Riemann-integrable on [0, 1].
Proof. Let P = {x0, . . . , xn} be any partition of [0.1]. Since both Qand R \ Q are dense in R, for each 1 †i †n, âqi â Q, ti 6â Q such thatqi, ti â [xiâ1, xi].
But f(qi) = 0 and f(ti) = 1, so that mi = 0 , Mi = 1 for all 1 †i †n.This implies that L(P ; f) = 0 and U(P ; f) = 1 for any partition P of I.Thus L(f) = 0 6= 1 = U(f) and f is not Riemann-integrable. ïżœ
P. Boily (uOttawa) 57
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
This last example underlines some of the shortcomings of the Riemannintegral.
The integral of this function should really be 0 on [0, 1]: the set R \Q is somuch larger than Q that whatever happens on Q should largely be irrelevant.
There are various theories of integration â as we shall see in a laterchapter, the Lebesgue-Borel integral of f on [0, 1] is indeed 0.
Other issues arise with the Riemann integral, which we will discuss inthe coming sections.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5.2.1 â Riemann Criterion
For the time being, we will focus on two fundamental questions associatedwith the Riemann integral of a function over an interval [a, b]: does itexist? If so, what value does it take?
But the direct approach is cumbersome, even for the simplest of functions.
The following result allows us to bypass the need to compute L(f) andU(f) to determine if a function is Riemann-integrable or not.
Theorem 52. (Riemannâs Criterion)Let I = [a, b] and f : I â R be a bounded function. Then f is Riemann-integrable if and only if âΔ > 0, âPΔ a partition of I such that the lower sumand the upper sum of f corresponding to PΔ satisfy U(PΔ; f)âL(PΔ; f) < Δ.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Proof. If f is Riemann-integrable, then L(f) = U(f) =â« baf .
Let Δ > 0. Sinceâ« baf â Δ
2 is not an upper bound of {L(P ; f) |P a partition of [a, b]}, there exists a partition P1 such that
â« b
a
f â Δ
2< L(P1; f) â€
â« b
a
f.
Using a similar argument, there exists a partition P2 such that
â« b
a
f +Δ
2â„ U(P2; f) >
â« b
a
f.
Set PΔ = P1 âȘ P2. Then PΔ is a refinement of P1 and P2.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Consequently,
â« b
a
f â Δ
2< L(P1; f) †L(PΔ; f) †U(PΔ; f) †U(P2; f) <
â« b
a
f +Δ
2
which implies thatU(PΔ; f)â L(PΔ; f) < Δ.
Conversely, let Δ > 0 and PΔ be such that U(PΔ; f) â L(PΔ; f) < Δ. SinceU(f) †U(PΔ; f) and L(f) â„ L(PΔ; f), then
0 †U(f)â L(f) †U(PΔ; f)â L(PΔ; f) < Δ.
But Δ > 0 was arbitrary, so U(f) â L(f) = 0, which in turns implies thatU(f) = L(f) and that f is Riemann-integrable on [a, b]. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
The Riemann Criterion is illustrated below for a continuous function:
The smaller the shaded area is, the closer U(P ; f) and L(P ; f) are toâ« baf .
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
There are 2 instances where the Riemann-integrability of a function f on[a, b] is guaranteed: when f is monotone, and when it is continuous.
Theorem 53. Let I = [a, b] and f : I â R be a monotone function on I.Then f is Riemann-integrable on I.
Proof. We show that the result holds for increasing functions. The prooffor decreasing functions is similar. Let
Pn = {xi = a+ i(bâan
): i = 0, . . . , n}
be the partition of I into n equal sub-intervals. Since f is increasing on I,we have, for 1 †i †n,
mi= inf{f(x) | x â [xiâ1, xi]} = f(xiâ1),
Mi = sup{f(x) | x â [xiâ1, xi]} = f(xi).
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Hence,
U(Pn; f)â L(Pn; f)=nâi=1
Mi(xi â xiâ1)ânâi=1
mi(xi â xiâ1)
=
nâi=1
(Mi âmi)(xi â xiâ1)
=bâ an
nâi=1
(f(xi)â f(xiâ1))
=bâ an
[f(x1)â f(x0) + · · ·+ f(xn)â f(xnâ1)
]=bâ an
(f(b)â f(a)) â„ 0.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Let Δ > 0. By the Archimedean Property, âNΔ â N such that
(bâ a)(f(b)â f(a))Δ
< n.
Set PΔ = Pn. Then
U(PΔ; f)â L(PΔ; f) <bâ aNΔ
(f(b)â f(a)) < Δ,
and f is Riemann-integrable on [a, b], according to Riemannâs Criterion. ïżœ
Theorem 54. Let I = [a, b] and f : I â R be continuous, with a < b.Then f is Riemann-integrable on I.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Proof. Let Δ > 0.
According to Theorem 38, f is uniformly continuous on I. Hence âΎΔ > 0s.t. |f(x)â f(y)| < Δ
bâa whenever |xâ y| < ΎΔ and x, y â [a, b].
Pick n â N such that bâan < ΎΔ and let
PΔ = {xi = a+ i(bâan
): i = 0, . . . , n}
be the partition of [a, b] into n equal sub-intervals.
As f is continuous on [xiâ1, xi], âui, vi â [xiâ1, xi] such that
mi= inf{f(x) | x â [xiâ1, xi]} = f(ui),
Mi= sup{f(x) | x â [xiâ1, xi]} = f(vi),
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
for all 1 †i †n, according to the Max/Min Theorem. (Note that|ui â vi| †bâa
n < ΎΔ for all i.)
Hence,
U(PΔ; f)â L(PΔ; f) =nâi=1
(Mi âmi)(xi â xiâ1) =bâ an
nâi=1
(f(vi)â f(ui))
<bâ an
nâi=1
Δ
bâ a= Δ,
by uniform continuity of f .
According to Riemannâs Criterion, f is thus Riemann-integrable. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5.2.2 â Properties of the Riemann Integral
The Riemann integral has a whole slew of interesting properties.
Theorem 55. (Properties of the Riemann Integral)Let I = [a, b] and f, g : I â R be Riemann-integrable on I. Then
(a) f + g is Riemann-integrable on I, withâ« ba(f + g) =
â« baf +
â« bag;
(b) if k â R, k · f is Riemann-integrable on I, withâ« bak · f = k
â« baf ;
(c) if f(x) †g(x) âx â I, thenâ« baf â€
â« bag, and
(d) if |f(x)| †K âx â I, thenâŁâŁâŁâ« ba f âŁâŁâŁ †K(bâ a).
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Proof. We use a variety of pre-existing results.
(a) Let Δ > 0. Since f, g are Riemann-integrable, âP1, P2 partitions of Isuch that U(P1; f)â L(P1; f) <
Δ2 and U(P2; g)â L(P2; g) <
Δ2.
Set P = P1 âȘ P2. Then P is a refinement of P1 and P2, and
U(P ; f + g) †U(P ; f) + U(P ; g)
< L(P ; f) + L(P ; g) + Δ †L(P ; f + g) + Δ, (1)
since, over non-empty subsets of I, we have
inf{f(x) + g(x)}℠inf{f(x)}+ inf{g(x)}sup{f(x) + g(x)}†sup{f(x)}+ sup{g(x)}.
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Hence f + g is Riemann-integrable according to Riemannâs Criterion.Furthermore, we see from (1) thatâ« b
a
(f + g) †U(P ; f + g) < L(P ; f) + L(P ; g) + Δ â€â« b
a
f +
â« b
a
g + Δ
andâ« b
a
f +
â« b
a
g †U(P ; f) + U(P ; g) < L(P ; f + g) + Δ â€â« b
a
(f + g) + Δ.
Since Δ > 0 is arbitrary,⫠baf +
â« bag â€
â« ba(f + g) â€
â« baf +
â« bag, from
which we conclude thatâ« ba(f + g) =
â« baf +
â« bag.
(b) The proof for k = 0 is trivial. We show that the result holds for k < 0(the proof for k > 0 is similar).
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Let P = {x0, . . . , xn} be a partition of I. Since k < 0, we haveinf{kf(x)} = k sup{f(x)} over non-empty subsets of I, and so we haveL(P ; kf) = kU(P ; f). In particular,
L(kf) = sup{L(P ; kf) | P a partition of I}= sup{kU(P ; f) | P a partition of I}= k inf{U(P ; f) | P a partition of I} = kU(f)
Similarly, U(P ; kf) = kL(P ; f) and U(fk) = kL(f), so
L(fk) = kU(f) = kL(f)ïžž ïž·ïž· ïžžsince f is R-int.
= U(kf).
Thus kf is Riemann-integrable on I andâ« bakf = L(k) = kU(f) =
â« baf.
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(c) We start by showing that if h : I â R is integrable on I and h(x) â„ 0
for all x â I, thenâ« bah(x) â„ 0.
Let P0 = {a, b} = {x0, x1} and m1 = inf{h(x) | x â [a, b]} â„ 0.Then,
0 †m1(bâ a) = L(P0;h) †L(P ;h)for any partition P of I, as P â P0. But h is Riemann-integrable byassumption, thus
â« b
a
h = sup{L(P ;h) | P a partition of I} â„ L(P0;h) â„ 0.
Then, set h = g â f .
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By hypothesis, h(x) = g(x)â f(x) â„ 0. Thenâ« b
a
h =
â« b
a
(g â f) =â« b
a
g ââ« b
a
f â„ 0,
which implies thatâ« bag â„
â« baf .
(d) Let P0 = {a, b} = {x0, x1}. As always, set m1 = inf{f(x) | x â [a, b]},and M1 = sup{f(x) | x â [a, b]}. Then for any partition P of I, wehave
m1(bâ a)= L(P0; f) †L(P ; f) †L(f) =â« b
a
f
= U(f) †U(P ; f) †U(P0; f) =M1(bâ a).
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In particular,
m1(bâ a) â€â« b
a
f â€M1(bâ a).
Now, if |f(x)| †K for all x â I, then âK †m1 and M1 †K so that
âK(bâ a) †m1(bâ a) â€â« b
a
f â€M1(bâ a) †K(bâ a),
so that |â« baf | †K(bâ a). ïżœ
When all the functions involved are non-negative, these results and the nextone are compatible with the area under the curve from calculus (more onthis in the next section).
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Theorem 56. (Additivity of the Riemann Integral)Let I = [a, b], c â (a, b), and f : I â R be bounded on I. Then f isRiemann-integrable on I if and only if it is Riemann-integrable on I1 = [a, c]
and on I2 = [c, b]. When that is the case,â« baf =
â« caf +
â« bcf .
Proof. We start by assuming that f is Riemann-integrable on I.
Let Δ > 0. According to the Riemann Criterion, âPΔ a partition of Isuch that U(PΔ; f) â L(PΔ; f) < Δ. Now, set P = PΔ âȘ {c}. Then P is arefinement of PΔ so that
U(P ; f)â L(P ; f) †U(PΔ; f)â L(PΔ; f) < Δ.
Set P1 = P â© I1 and P2 = P â© I2.
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Then Pi is a partition of Ii, and
Δ > U(P ; f)â L(P ; f) â„ U(P1; f) + U(P2; f)â L(P1; f)â L(P2; f)
=[U(P1; f)â L(P1; f)
]+[U(P2; f)â L(P2; f)
]Consequently, U(Pi; f) â L(Pi; f) < Δ for i = 1, 2 and f is Riemann-integrable on I1 and I2, according to the Riemann Criterion.
Now assume that f is Riemann-integrable on I1 and I2.
Let Δ > 0. According to the Riemann Criterion, for i = 1, 2, âPi apartition of Ii such that
U(Pi; f) + L(Pi; f) <Δ
2.
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Set P = P1 âȘ P2. Then P is a partition of I. Furthermore,
U(P ; f)â L(P ; f) = U(P1; f) + U(P2; f)â L(P1; f)â L(P2; f)
= U(P1; f)â L(P1; f) + U(P2; f)â L(P2; f) <Δ
2+Δ
2= Δ,
thus f is Riemann-integrable on I according the Riemann Criterion.
Finally, letâs assume that f is Riemann-integrable on I (and so on I1, I2),or vice-versa.
Let P1, P2 be partitions of I1, I2, respectively, such that
U(Pi; f)â L(Pi; f) <Δ
2, i = 1, 2.
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Set P = P1 âȘ P2. Then P is a partition of I and
â« b
a
f †U(P ; f) = U(P1; f) + U(P2; f)
< L(P1; f) + L(P2; f) + Δ =
â« c
a
f +
â« b
c
f + Δ.
Similarly,
â« b
a
f â„ L(P1; f) + L(P2; f) > U(P1; f) + U(P2; f)â Δ â„â« c
a
f +
â« b
c
f â Δ
Since Δ > 0 is arbitrary,⫠baf =
â« caf +
â« bcf . ïżœ
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Theorem 57. (Composition Theorem for Integrals)Let I = [a, b] and J = [α, ÎČ], f : I â R Riemann-integrable on I,Ï : J â R continuous on J and f(I) â J . Then Ï âŠ f : I â R isRiemann-integrable on I.
Proof. Let Δ > 0, K = sup{|Ï(x)| | x â J} (guaranteed to exist by theMax/Min theorem) and ΔâČ = Δ
bâa+2K .
Since Ï is uniformly continuous on J (being continuous on a closed,bounded interval), âΎΔ > 0 s.t.
|xâ y| < ΎΔ, x, y,â J =â |Ï(x)â Ï(y)| < ΔâČ.
Without loss of generality, pick ΎΔ < ΔâČ.
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Since f is Riemann-integrable on I, âP = {x0, . . . , xn} a partition ofI = [a, b] s.t.
U(P ; f)â L(P ; f) < ÎŽ2Δ
(according to Riemannâs criterion).
We show that U(P ;Ï âŠ f) â L(P ;Ï âŠ f) < Δ, and so that Ï âŠ f isRiemann-integrable according to Riemannâs criterion.
on [xiâ1, xi] for i = 1, . . . , n, set
mi = inf{f(x)}, Mi = sup{f(x)}, mi = inf{Ï(f(x))}, Mi = sup{Ï(f(x))}.
With those, set A = {i |Mi âmi < ΎΔ}, B = {i |Mi âmi ℠ΎΔ}.
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If i â A, then
x, y â [xiâ1, xi] =â |f(x)â f(y)| â€Mi âmi < ΎΔ,
so |Ï(f(x))âÏ(f(y)| < ΔâČ âx, y â [xiâ1, xi]. In particular, Miâmi †ΔâČ.
If i â B, then
x, y â [xiâ1, xi] =â |Ï(f(x))âÏ(f(y))| †|Ï(f(x))|+|Ï(f(y))| †2K.
In particular, Mi â mi †2K, since âK †mi †Ï(z) †Mi †K for allz â [xiâ1, xi].
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Then
U(P ;Ï âŠ f)âL(P ;Ï âŠ f) =nâi=1
(Mi â mi)(xi â xiâ1)
=âiâA
(Mi â mi)(xi â xiâ1) +âiâB
(Mi â mi)(xi â xiâ1)
†ΔâČâiâA
(xi â xiâ1) + 2KâiâB
(xi â xiâ1)
†ΔâČ(bâ a) + 2KâiâB
(Mi âmi)
ΎΔ(xi â xiâ1)
ΔâČ(bâ a) + 2K
ΎΔ
nâi=1
(Mi âmi)(xi â xiâ1).
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By earlier work in the proof, we have
nâi=1
(Mi âmi)(xi â xiâ1) †U(P ; f)â L(P ; f) < ÎŽ2Δ,
so that
U(P ;Ï âŠ f)â L(P ;Ï âŠ f) < ΔâČ(bâ a) + 2K
ΎΔ· Ύ2Δ
= ΔâČ(bâ a) + 2KΎΔ < ΔâČ(bâ a) + 2KΔâČ
= ΔâČ(bâ a+ 2K) = Δ,
which completes the proof. ïżœ
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The proof of Composition Theorem requires the intervals I and J to beclosed, as the following example shows.
Example: Let f, Ï : (0, 1) â R be defined by f(x) = x and Ï(x) = 1x.
Then f is Riemann-integrable on (0, 1), Ï is continuous on (0, 1), butÏ âŠ f : (0, 1)â R, (Ï âŠ f)(x) = 1/x, is not Riemann-integrable on (0, 1).
Note, however, that there are examples of functions defined on openintervals for which the conclusion of the Composition Theorem holds.
Example: Let f, Ï : (0, 1) â R be defined by f(x) = x and Ï(x) = x.Then f is Riemann-integrable on (0, 1), Ï is continuous on (0, 1), andÏ âŠ f : (0, 1)â R, (Ï âŠ f)(x) = x, is Riemann-integrable on (0, 1).
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The Composition Theorem can be used to show a variety of results.
Theorem 58. Let I = [a, b] and f, g : I â R be Riemann-integrable on I.
Then fg and |f | are Riemann-integrable on I, andâŁâŁâŁâ« ba f âŁâŁâŁ †⫠ba |f |.
Proof. The function defined by Ï(t) = t2 is continuous. By the CompositionTheorem, Ï âŠ (f + g) = (f + g)2 and Ï âŠ (f â g) = (f â g)2 are bothRiemann-integrable on I.
But the product fg can be re-written as
fg =1
4
[(f + g)2 â (f â g)2
].
According to Theorem 55, fg is Riemann-integrable on I.
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Now, consider the function defined by Ï(t) = |t|. It is continuous, soÏ âŠ f = |f | is R-integrable on I according to the Composition Theorem.
Pick c â {±1} such that câ« baf â„ 0. HenceâŁâŁâŁâŁâŁ
â« b
a
f
âŁâŁâŁâŁâŁ = c
â« b
a
f =
â« b
a
cf â€â« b
a
|f |,
since cf(x) †|f(x)| for all x â I. ïżœ
Note that even though the product of Riemann-integrable functions is
itself Riemann-integrable there is no simple way to expressâ« bafg in terms
ofâ« baf and
â« bag.
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A surprising result is that the composition of Riemann-integrable functionsneed not be Riemann-integrable. A counter example is provided below.
Example: Let I = [0, 1] and let f : I â R be Thomaeâs function:
f(x) =
1, x = 0
1/n, x = m/n â Q, gcd(m,n) = 1
0, x 6â Q
It can be shown that f is Riemann-integrable on [0, 1] and thatâ« 1
0f = 0.
Consider the function g : [0, 1] â R defined by g(x) ⥠1 on (0, 1] and
g(0) = 0. Then g is Riemann-integrable on [0, 1], withâ« 1
0g = 1, but
g ⊠f : [0, 1]â R is the Dirichlet function, and is therefore not Riemann-integrable on [0, 1].
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5.2.3 â Fundamental Theorem of Calculus
With Descartesâ creation of analytical geometry, it became possible to findthe tangents to curves that could be described algebraically.
Fermat then showed the connection between that problem and the problemof finding the maximum/minimum of a (continuous) function.
Newton and Leibniz, in the 1680s, then discovered that computing thearea underneath a curve is exactly the opposite of finding the tangent.
Calculus provided a general framework to solve problems that had hitertobeen very difficult to solve (and even then, only in specific circumstance).
In this section, we study the connection between these concepts.
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Theorem 59. (Fundamental Theorem of Calculus, 1st version)Let I = [a, b], f : I â R be Riemann-integrable on I, and F : I â R besuch that F is continuous on I and differentiable on (a, b). If F âČ(x) = f(x)
for all x â (a, b), thenâ« baf = F (b)â F (a).
Proof. Let Δ > 0. Since f is Riemann-integrable on I, âPΔ a partition of Isuch that
U(PΔ; f)â L(PΔ; f) < Δ.
Applying the Mean Value Theorem to F on [xiâ1, xi] for each 1 †i †n,we conclude that âti â (xiâ1, xi) such that
F (xi)â F (xiâ1)xi â xiâ1
= F âČ(ti) = f(ti), 1 †i †n.
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Let
mi = inf{f(x) | x â [xiâ1, xi]}, Mi = sup{f(x) | x â [xiâ1, xi]}
for 1 †i †n. Then
L(PΔ; f) =
nâi=1
mi(xi â xiâ1) â€nâi=1
f(ti)(xi â xiâ1)
=nâi=1
(F (xi)â F (xiâ1)) = F (b)â F (a),
and, similarly, U(PΔ; f) â„ F (b)â F (a).
Then L(PΔ; f) †F (b)â F (a) †U(PΔ; f) for all Δ > 0.
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Since we have L(PΔ; f) â€â« baf †U(PΔ; f) and U(PΔ; f) â L(PΔ; f) < Δ,
for all Δ > 0, we must also haveâŁâŁâŁâŁâŁâ« b
a
f â (F (b)â F (a))
âŁâŁâŁâŁâŁ < Δ, for all Δ > 0,
so thatâ« baf = F (b)â F (a). ïżœ
This classical calculus result is quite useful in applications.
We will see in a later chapter that the Fundamental Theorem of Calculus(1st version) is a special case of the general result known as StokesâTheorem.
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Theorem 60. (Fundamental Theorem of Calculus, 2nd version)Let I = [a, b], f : I â R be Riemann-integrable on I, and define a functionF : I â R by F (x) =
â« xaf . Then F is continuous on I. Furthermore, if f
is continuous at c â (a, b), then F is differentiable at c and F âČ(c) = f(c).
Proof. Since f is Riemann-integrable on I, then f is bounded on I. LetK > 0 be such that |f(x)| < K for all x â I.
Let x â I and Δ > 0. Set ΎΔ =ΔK . Then whenever |x â y| < ΎΔ =
ΔK and
y â I, we have
|F (y)â F (x)| =âŁâŁâŁâŁâ« y
a
f ââ« x
a
f
âŁâŁâŁâŁ = âŁâŁâŁâŁâ« y
x
f
âŁâŁâŁâŁ †K|xâ y| < Δ.
Then F is uniformly continuous on I, and so is continuous on I.
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Now assume that f is continuous at c and let Δ > 0. Then âΎΔ > 0 suchthat |f(x)â f(c)| < Δ whenever |xâ c| < ΎΔ and x â I.
Thus, if 0 †|h| = |xâ c| < ΎΔ and x â I, we have
âŁâŁâŁâŁF (c+ h)â F (c)h
â f(c)âŁâŁâŁâŁ =
âŁâŁâŁâŁâŁ1hâ« c+h
a
f â 1
h
â« c
a
f â f(c)
âŁâŁâŁâŁâŁ=
âŁâŁâŁâŁâŁ1hâ« c+h
c
f â 1
h
â« c+h
c
f(c)
âŁâŁâŁâŁâŁ =âŁâŁâŁâŁâŁ1hâ« c+h
c
(f â f(c))
âŁâŁâŁâŁâŁâ€ 1
|h|
âŁâŁâŁâŁâŁâ« c+h
c
(f â f(c))
âŁâŁâŁâŁâŁ < 1
|h|· Δ
âŁâŁâŁâŁâŁâ« c+h
c
1
âŁâŁâŁâŁâŁ = 1
|h|· Δ|h| = Δ,
which is to say, F âČ(c) = f(c). ïżœ
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The first version of the Fundamental Theorem Calculus provides ajustification of the method used to evaluate definite integrals in calculus;the second version, which allows the upper bound of the Riemann integralto vary, provides a basis for finding antiderivatives.
Let I = [a, b] an f : I â R. An antiderivative of f on I is a differentiablefunction F : I â R such that F âČ(x) = f(x) for all x â I.
If f is Riemann-integrable on I, the function F : I â R defined byF (x) =
â« xaf for x â I is the indefinite integral of f on I.
If f is Riemann-integrable on I and if F is an antiderivative of f on I, then
â« b
a
f = F (b)â F (a).
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However, Riemann-integrable functions on I may not have antiderivativeson I (such as the signum and Thomaeâs functions), and functions withantiderivatives may not be Riemann-integrable on I (such as the reciprocalof the square root function on [0, 1]).
If f is Riemann-integrable on I, then F (x) =â« xaf exists. Moreover,
if f is continuous on I, than F is an antiderivative of f on I, sinceF âČ(x) = f(x) for all x â I.
Continuous functions thus always have antiderivatives (even if they canâtbe expressed using elementary functions).
But if f is not continuous on I, the indefinite integral F may not bean antiderivative of f on I â it may fail to be differentiable at certain pointsof I, or F âČ may exists but be different from f at various points of I.
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5.2.4 â Evaluation of Integrals
We complete this chapter by presenting some common methods used toevaluate integrals.
We provide the proof of two well-known calculus results.
Theorem 61. (Integration by Parts)Let f, g : [a, b] â R both be Riemann-integrable on [a, b], withantiderivatives F,G : [a, b]â R, respectively. Then
â« b
a
Fg = F (b)G(b)â F (a)G(a)ââ« b
a
fG.
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Proof. Let H : [a, b] â R be defined by H = FG. As F and G are bothdifferentiable, so is H: H âČ = F âČG+ FGâČ = fG+ Fg.
Thenâ« baH âČ = H(b)âH(a), soâ« b
a
(fG+Fg) = F (b)G(b)âF (a)G(a) =ââ« b
a
Fg = H(b)âH(a)ââ« b
a
fG.
This completes the proof. ïżœ
Theorem 62. (First Substitution Theorem)Let J = [α, ÎČ], and Ï â R be a function with a continuous derivative onJ . If f : I â R is continuous on I = [a, b] â Ï(J), thenâ« ÎČ
α
(f ⊠Ï)ÏâČ =â« Ï(ÎČ)
Ï(α)
f.
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Proof. Since f is continuous on I, it is Riemann-integrable on I and so wecan define a function F : I â R through
F (x) =
â« x
Ï(α)
f, x â I.
By construction F is continuous and differentiable on I. Furthermore,F âČ = f on I, according to the second version of the Fundamental Theoremof Calculus.
Define H : J â R by H = F ⊠Ï. Then H is differentiable onI, being the composition of two differentiable functions on I, andH âČ = (F âČ âŠ Ï)ÏâČ = (f âČ âŠ Ï)ÏâČ is Riemann-integrable since Ï, f âŠ Ï areRiemann-integrable (being continuous) on I, according to Theorem 58.
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The first version of the Fundamental Theorem of Calculus then yields
â« ÎČ
α
(f ⊠Ï)ÏâČ =â« ÎČ
α
H âČ = H(ÎČ)âH(α) = F (Ï(ÎČ))â F (Ï(α)) =â« Ï(ÎČ)
Ï(α)
f,
which completes the proof. ïżœ
Theorem 63. (Second Substitution Theorem)Let J = [α, ÎČ], and Ï â R be a function with a continuous derivative onJ and such that ÏâČ 6= 0 on J . Let I = [a, b] â Ï(J), and Ï : I â R be theinverse of Ï (which exists as Ï is montoone). If f : I â R is continuouson I, then â« ÎČ
α
f âŠ Ï =
â« Ï(ÎČ)
Ï(α)
fÏâČ.
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Theorem 64. (Mean Value Theorem for Integrals)Let I = [a, b], f : I â R be continuous on I, and p : I â R beRiemann-integrable on I, with p â„ 0 on I. Then âc â (a, b) such that
â« b
a
fp = f(c)
â« b
a
p.
Theorem 65. (Squeeze Theorem for Integrals)Let I = [a, b] and f †g †h : I â R be bounded on I. If f, h are
Riemann-integrable on I withâ« baf =
â« bah, then g is Reimann-integrable on
I andâ« bag =
â« baf =
â« bah.
The proofs of these three last theorems are left as an exercise.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5.3 â Exercises
1. Use the definition to find the derivative of the function defined by g(x) = 1x, x â R,
x 6= 0.
2. Prove that the derivative of an even differentiable function is odd, and vice-versa.
3. Let a > b > 0 and n â N with n â„ 2. Show that a1/n â b1/n < (aâ b)1/n.
4. Let f : [a, b]â R be continuous on [a, b] and differentiable on (a, b). Show that if
limxâa
fâČ(x) = A, then f âČ(a) exists and equals A.
5. If x > 0, show 1 + 12xâ
18x
2 â€â1 + x †1 + 1
2x.
6. Let a â R and f : Râ R be defined by
f(x) =
{x2 if x â„ 0,
ax if x < 0.
For which values of a is f differentiable at x = 0? For which values of a is f
continuous at x = 0?
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
7. Let f : [a, b]â R be continuous on [a, b] and differentiable on (a, b). Show that f
is Lipschitz if and only if f âČ is bounded on (a, b).
8. Prove thatâ« 1
0g = 1
2 if
g(x) =
{1 x â (12, 1]
0 x â [0, 12].
Is that still true if g(12) = 7 instead?
9. Let f : [a, b]â R be bounded and s.t. f(x) â„ 0 âx â [a, b]. Show L(f) â„ 0.
10. Let f : [a, b]â R be increasing on [a, b]. If Pn partitions [a, b] into n equal parts,
show that
0 †U(Pn; f)ââ« b
a
f â€f(b)â f(a)
n(bâ a).
11. Let f : [a, b]â R be an integrable function and let Δ > 0. If PΔ is the partition whose
existence is asserted by the Riemann Criterion, show that U(P ; f) â L(P ; f) < Δ
for all refinement P of PΔ.
12. Let a > 0 and J = [âa, a]. Let f : J â R be bounded and let Pâ be the set of
symmetric partitions of J that contain 0. Show L(f) = sup{L(P ; f) : P â Pâ}.
P. Boily (uOttawa) 102
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
13. Let a > 0 and J = [âa, a]. Let f be integrable on J . If f is even (i.e.
f(âx) = f(x) for all x), show thatâ« a
âaf = 2
â« a
0
f.
If f is odd (i.e. f(âx) = âf(x) for all x), show that
â« a
âaf = 0.
14. Give an example of a function f : [0, 1]â R that is not integrable on [0, 1], but s.t.
|f | is integrable on [0, 1].
15. Let f : [a, b] â R be integrable on [a, b]. Show |f | is integrable on [a, b] directly
(without using a result seen in class).
16. If f is integrable on [a, b] and 0 †m †f(x) â€M for all x â [a, b], show that
m â€[
1
bâ a
â« b
a
f2
]1/2
â€M.
P. Boily (uOttawa) 103
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
17. If f is continuous on [a, b] and f(x) â„ 0 for all x â [a, b], show there exists
c â [a, b] s.t.
f(c) =
[1
bâ a
â« b
a
f2
]1/2
.
18. If f is continuous on [a, b] and f(x) > 0 for all x â [a, b], show that 1f is integrable
on [a, b].
19. Let f be continuous on [a, b]. Define H : [a, b]â R by
H(x) =
â« b
x
f for all x â [a, b].
Find H âČ(x) for all x â [a, b].
20. Suppose f : [0,â)â R is continuous and f(x) 6= 0 for all x > 0. If
(f(x))2= 2
â« x
0
f for all x > 0,
show that f(x) = x for all x â„ 0.
P. Boily (uOttawa) 104
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
21. Let f, g : [a, b]â R be continuous and s.t.â« b
a
f =
â« b
a
g.
Show that there exists c â [a, b] s.t. f(c) = g(c).
22. If f : [0, 1]â R is continuous andâ« x0f =
â« 1
xf for all x â [0, 1], show that f ⥠0.
23. Let f : [0, 3]â R be defined by
f(x) =
x x â [0, 1)
1 x â [1, 2)
x x â [2, 3]
.
Find F : [0, 3]â R, where
F (x) =
â« x
0
f.
Where is F differentiable? What is F âČ there?
P. Boily (uOttawa) 105
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
24. Let f : [a, b]â R be continuous, f â„ 0 on [a, b], andâ« baf = 0. Show that f ⥠0
on [a, b].
25. Let f : [a, b] â R be continuous and letâ« baf = 0. Show âc â [a, b] such that
f(c) = 0.
26. Computed
dx
â« x
âxet2dt.
27. Let f : [a, b]â R be Riemann-integrable on [a+ ÎŽ, b] and unbounded in the interval
(a, a+ ÎŽ) for every 0 < ÎŽ < bâ a. Defineâ« b
a
f = limÎŽâ0+
â« b
a+ÎŽ
f,
where ÎŽ â 0+ means that ÎŽ â 0 and ÎŽ > 0. A similar construction allows us to
define â« b
a
g = limÎŽâ0+
â« bâÎŽ
a
g.
Such integrals are said to be improper; when the limits exist, they are further said to
be convergent.
P. Boily (uOttawa) 106
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
How can the expression â« 1
0
1â|x|
dx
be interpreted as an improper integral? Is it convergent? If so, what is its value?
28. For which values of s does the integralâ« 1
0xs dx converge?
P. Boily (uOttawa) 107
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Solutions
1. Proof. From calculus, we âknowâ that gâČ(x) = â 1x2
.
Let c â R s.t. c 6= 0. Set ac = c2 and bc = 3c
2 . Clearly, if c > 0,0 < ac < c < bc, whereas bc < c < ac < 0 if c < 0. In both cases,1|x| â€
1|ac| whenever x lies between ac and bc. We restrict g on the
interval between ac and bc (denote this interval by A).
Let Δ > 0 and set ΎΔ = |ac|c2Δ. Then whenever 0 < |x â c| < ΎΔ andx â A, we haveâŁâŁâŁâŁâŁ 1x â 1
c
xâ c+
1
c2
âŁâŁâŁâŁâŁ =âŁâŁâŁâŁ câ xxc(xâ c)
+1
c2
âŁâŁâŁâŁ = âŁâŁâŁâŁ 1c2 â 1
xc
âŁâŁâŁâŁ = |xâ c||x|c2†|xâ c||ac|c2
<ΎΔ|ac|c2
= Δ,
which validates our calculus guess. ïżœ
P. Boily (uOttawa) 108
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
2. Proof. If f is even, then f(x) = f(âx) for all x â R. Let g(x) = f(âx).Then g is differentiable by the chain rule and f(x) = g(x) for all x â R.
Furthermore,
f âČ(x) = gâČ(x) = (f(âx))âČ = f âČ(âx) · â1,
that is, âf âČ(âx) = f âČ(âx), or f âČ is odd. The other statement is provedsimilarly. ïżœ
P. Boily (uOttawa) 109
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
3. Proof. Consider the continuous function f : [1,â) â R defined byf(x) = x1/n â (xâ 1)1/n, whose derivative is
f âČ(x) =1
n
(x
1ânn â (xâ 1)
1ânn
).
Now,
0 †xâ 1 < x, âx â„ 1 =â 0 †(xâ 1)n < xn, âx â„ 1, n â„ 2
⎠0 †(xâ 1)nnâ1 < x
nnâ1, âx â„ 1, n â„ 2
and so1
xnnâ1
<1
(xâ 1)nnâ1
,
or x1ânn < (xâ 1)
1ânn for all x â„ 1, n â„ 2.
P. Boily (uOttawa) 110
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Hence f âČ(x) < 0 for all x â [1,â), that is f is strictly decreasing over[1,â). But f(ab) < f(1), as a
b > 1. But
f(ab
)=(ab
)1n â
(abâ 1)1n=
1
b1n
(a
1n â (aâ b)1n
)and f(1) = 1, so
1
b1n
(a
1n â (aâ b)1n
)< 1,
that is a1n â (aâ b)1n < b
1n, which completes the proof. ïżœ
P. Boily (uOttawa) 111
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
4. Proof. Let x â (a, b). By the Mean Value Theorem, âcx â (a, x) s.t.
f(x)â f(a)xâ a
= f âČ(cx).
When x â a, cx â a (indeed, let Δ > 0 and set ΎΔ = Δ; then|cx â a| < |xâ a| < ΎΔ = Δ whenever 0 < |xâ a| < ΎΔ). Then
limxâa
f âČ(cx) = limcxâa
f âČ(cx) = A
by hypothesis. Hence limxâa
f âČ(x) exists and so
f âČ(a) = limxâa
f(x)â f(a)xâ a
= limxâa
f âČ(x) = A
exists. ïżœ
P. Boily (uOttawa) 112
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
5. Proof. Let x0 = 0 and f(x) =â1 + x. According to Taylorâs
theorem, since f is C3 when x > 0, f(x) = P1(x) + R1(x) andf(x) = P2(x) +R2(x), where
P1(x) = f(x0) + f âČ(x0)(xâ x0) =â1 + 0 +
1
2â1 + 0
x = 1 +1
2x
P2(x) = P1(x) +f âČâČ(x0)
2(xâ x0)2 = 1 +
1
2xâ 1
8 3â1 + 0
x2 = 1 +1
2xâ 1
8x2
R1(x) =f âČâČ(c1)
2(xâ x0)2 = â
1
8 3â1 + c1
x2, for some c1 â [0, x]
R2(x) =f âČâČâČ(c2)
6(xâ x0)3 =
3
48 5â1 + c2
x3, for some c2 â [0, x].
When x > 0, R1(x) †0 and R2(x) â„ 0, so P2(x) †f(x) †P1(x). ïżœ
P. Boily (uOttawa) 113
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
6. Solution. We have
f âČ+(0) = limxâ0+
f(x)â f(0)xâ 0
= limxâ0+
x2
x= limxâ0+
x = 0
and
f âČâ(0) = limxâ0â
f(x)â f(0)xâ 0
= limxâ0â
ax
x= limxâ0+
a = a.
Thus, f is differentiable at x = 0 if and only if a = 0.
Since both x2 and ax are continuous functions, we have
limxâ0+
f(x) = limxâ0+
x2 = 0 = f(0) = 0 = limxâ0â
ax = limxâ0â
f(x)
and the the function f is continuous at x = 0 for all values of a. ïżœ
P. Boily (uOttawa) 114
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
7. Proof. Suppose that f satisfies the Lipschitz condition on [a, b] withconstant M . Then, for all x0 â (a, b), we haveâŁâŁâŁâŁf(x)â f(x0)xâ x0
âŁâŁâŁâŁ â€M âx â (a, b) \ {x0}.
Thus
|f âČ(x0)| =âŁâŁâŁâŁ limxâx0
f(x)â f(x0)xâ x0
âŁâŁâŁâŁ = limxâx0
âŁâŁâŁâŁf(x)â f(x0)xâ x0
âŁâŁâŁâŁ â€M,
where we used the fact that the absolute value function is continuous topull the limit out of the absolute value. So the derivative of f is boundedon (a, b).
P. Boily (uOttawa) 115
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Now assume that |f âČ(x)| â€M for all x â (a, b). Let x, y â [a, b], x < y.Applying the Mean Value Theorem to f on the interval [x, y] yields theexistence of c â (x, y) such that
f(y)â f(x)y â x
= f âČ(c).
Thus âŁâŁâŁâŁf(x)â f(y)xâ y
âŁâŁâŁâŁ â€M =â |f(x)â f(y)| â€M |xâ y|.
This completes the proof. ïżœ
P. Boily (uOttawa) 116
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
8. Proof. Let Δ > 0 and define the partition PΔ = {0, 12â Δ,12+ Δ, 1}. Since
g is bounded on [0, 1], L(g) †U(g) exist and
L(g) ℠L(PΔ; g) =1
2â Δ and U(g) †U(PΔ; g) =
1
2+ Δ.
Hence1
2â Δ †L(g) †U(g) †1
2+ Δ, for all Δ > 0.
Since Δ > 0 is arbitrary, then 12 †L(g) †U(g) †1
2; by definition, g is
Riemann integrable on [0, 1] and L(g) = U(g) =â« baf = 1
2.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
If instead g(1/2) = 7, the exact same work as above yields
1
2â Δ †L(g) †U(g) †1
2+ 13Δ, for all Δ > 0.
Since Δ > 0 is arbitrary, then 12 †L(g) †U(g) †1
2; by definition, g is
also Riemann integrable on [0, 1] and L(g) = U(g) =â« baf = 1
2. ïżœ
P. Boily (uOttawa) 118
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
9. Proof. As f is bounded on [a, b], L(f) exists and the set
{f(x) : x â [a, b]} 6= â
is bounded below.
By completeness of R, m1 = inf{f(x) : x â [a, b]} exists.
Furthermore, m1 â„ 0 since f(x) â„ 0 for all x â [a, b].
Let P = {x0, x1} = {a, b} be the trivial partition of [a, b].
Then L(f) â„ L(P ; f) = m1(bâ a) â„ 0. ïżœ
P. Boily (uOttawa) 119
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
10. Proof. As f is increasing, it is monotone and thus Riemann integrableby a result seen in class (Theorem 53).
Then L(f) = U(f) =â« baf .
LetPn = {xi = a+ ibâan : i = 0, . . . , n}
be the partition of [a, b] into n equal sub-intervals.
By definition, L(Pn; f) â€â« baf and U(Pn; f) â„
â« baf . Then
U(Pn; f)â L(Pn; f) â„ U(Pn; f)ââ« b
a
f â„â« b
a
f ââ« b
a
f = 0.
In particular, U(Pn; f)ââ« baf â„ 0. As f is increasing on [a, b],
P. Boily (uOttawa) 120
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Mi = sup[xiâ1,xi]
{f(x)} = f(xi), mi = inf[xiâ1,xi]
{f(x)} = f(xiâ1), and
U(Pn; f)â L(Pn; f) =nâi=1
(Mi âmi)(xi â xiâ1)
=bâ an
nâi=1
(f(xi)â f(xiâ1)) =bâ an
(f(b)â f(a)).
Since L(Pn; f) â€â« baf , then
bâ an
(f(b)â f(a)) = U(Pn; f)âL(Pn; f) â„ U(Pn; f)ââ« b
a
f â„ 0. ïżœ
P. Boily (uOttawa) 121
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
11. Proof. Let P be a refinement of PΔ.
Then U(PΔ;f) ℠U(P ; f) and L(PΔ; f) †L(P ; f), and so
U(PΔ; f) ℠U(P ; f) ℠L(P ; f) ℠L(PΔ; f).
By the Riemann Criterion, U(PΔ; f) < Δ+ L(PΔ; f). Then
Δ+ L(P ; f) ℠Δ+ L(PΔ; f) > U(PΔ; f) ℠U(P ; f),
i.e. Δ+ L(P ; f) > U(P ; f), which completes the proof. ïżœ
P. Boily (uOttawa) 122
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
12. Proof. Let α = sup{L(P ; f) : P â Pâ}. By definition,
α †L(f) = sup{L(P ; f) : P is a partition of [âa, a]}.
Let Δ > 0 and PΔ = {x0, x1, . . . , xn} be a partition of [âa, a] s.t.L(f)â Δ < L(PΔ; f) †L(f). Such a partition exists as L(f)â Δ is notthe supremum of the aforementioned set.
Consider the set {0,±x0, . . . ,±xn}. Eliminate all the repetitions fromthis set and re-order its elements. Denote the new set by QΔ.
Then QΔ is a refinement of PΔ and QΔ â Pâ; so α â„ L(QΔ; f),and
L(f)â Δ < L(PΔ; f) †L(QΔ; f) †α †L(f),as Δ > 0 is arbitrary, L(f) = α. ïżœ
P. Boily (uOttawa) 123
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
13. Proof. As f is integrable over [âa, a], a result seen in class (Theorem56) implies that f is integrable over [0, a].
If f is even, let P â Pâ. Then there is a partition P of [0, a] s.t.L(P ; f) = 2L(P ; f) and vice-versa.
Indeed, letP = {xân, . . . , xâ1, x0, x1, . . . , xn},
where x0 = 0 and xâi = âxi for all i = 1, . . . , n.
Then P â Pâ.
Let mi = inf{f(x) : x â [xiâ1, xi]}, for i = ânâ 1, . . . , 0, . . . , n.
Since f is even, mi = mâi+1 for i = ânâ 1, . . . , 0, . . . , n.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Then
L(P ; f) =
0âi=ânâ1
mi(xi â xiâ1) +nâi=1
mi(xi â xiâ1)
= 2
nâi=1
mi(xi â xiâ1) = L(P ; f),
where P is a partition of [0, a]. This, combined with the previousexercise, yieldsâ« a
âaf = sup{L(P ; f) : P â Pâ} = sup{2L(P ; f) : P is a partition of [0, a]}
= 2 sup{L(P ; f) : P is a partition of [0, a]} = 2
â« a
0
f.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
If f is odd, consider the function h : Râ R given by
h(x) =
{1 if x â„ 0
â1 if x < 0.
The product fh is an even function, so
2
â« a
0
f = 2
â« a
0
hf =
â« a
âahf =
â« 0
âahf +
â« a
0
hf =
â« 0
âaâf +
â« a
0
f,
and soâ« a0f =
â« 0
âaâf = ââ« 0
âa f . Thenâ« a
âaf =
â« 0
âaf +
â« a
0
f = ââ« a
0
f +
â« a
0
f = 0. ïżœ
P. Boily (uOttawa) 126
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
14. Solution. Here is one example: f : [0, 1]â R, defined by f(x) = â1 ifx 6â Q and f(x) = 1 if x â Q.
The proof that f is not Riemann integrable is similar to the one done inclass.
How would you prove that |f | ⥠1 is Riemann integrable? (Hint:what do U(P ; f) and L(P ; f) look like?)
P. Boily (uOttawa) 127
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
15. Proof. Let Δ > 0.
By the Riemann Criterion, there exists a partition PΔ = {x0, . . . , xn} of[a, b] s.t. U(PΔ; f)â L(PΔ; f) < Δ.
For all i = 1, . . . , n, let
Mi = sup{f(x) : x â [xiâ1, xi]} and mi = inf{f(x) : x â [xiâ1, xi]}.
For all i = 1, . . . , n, then,
|f(x)â f(y)| â€Mi âmi on [xiâ1, xi].
As ||f(x)| â |f(y)|| †|f(x)â f(y)| †Mi âmi for all x, y â [xiâ1, xi],we have
Mi â mi â€Mi âmi,
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
where
Mi = sup{|f(x)| : x â [xiâ1, xi]} and mi = inf{|f(x)| : x â [xiâ1, xi]}
for all i = 1, . . . , n. Then
U(PΔ; |f |)â L(PΔ; |f |) =nâi=1
(Mi â mi
)(xi = xiâ1)
â€nâi=1
(Mi âmi) (xi = xiâ1) = U(PΔ; |f |)â L(PΔ; |f |) < Δ.
By the Riemann Criterion, this shows that |f | is integrable on [a, b]. ïżœ
P. Boily (uOttawa) 129
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
16. Proof. By hypothesis, m2 †f2(x) â€M2 for all x â [a, b].
As f is integrable on [a, b], so is f2, by a result seen in class (Theorem 58).
Then â« b
a
m2 â€â« b
a
f2 â€â« b
a
M2
by the âSqueeze Theorem for Integralsâ and so
m2(bâ a) â€â« b
a
f2 â€M2(bâ a).
We obtain the desired result by re-arranging the terms and extractingsquare roots. ïżœ
P. Boily (uOttawa) 130
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
17. Proof. By the Max/Min theorem, âx0, x1 â [a, b] s.t.
m = inf[a,b]{f(x)} = f(x0), M = sup
[a,b]
{f(x)} = f(x1).
Then by the preceding exercise, we have
f(x0) â€
[1
bâ a
â« b
a
f2
]1/2†f(x1).
As f is continuous on [x0, x1] (or [x1, x0]), the IVT states âc â [a, b] s.t.
f(c) =
[1
bâ a
â« b
a
f2
]1/2. ïżœ
P. Boily (uOttawa) 131
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
18. Proof. Since f is continous on [a, b] it is integrable on [a, b].
Since f is continuous and [a, b] is a closed bounded interval, thenf([a, b]) = [m,M ] is also closed bounded interval (a result seen in class).
Furthermore, 0 < m â€M since f(x) > 0 for all x â [a, b].
Let Ï : [m,M ]â R be defined by Ï(t) = 1t . Then Ï is continuous and
bounded on [m,M ] and so Ï âŠ f : [a, b]â R, defined by Ï(f(x)) = 1f(x)
is integrable on [a, b], by the Composition Theorem 57. ïżœ
P. Boily (uOttawa) 132
MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
19. Proof. Define F (x) =â« xaf .
Since f is continuous, F is differentiable and F âČ(x) = f(x) for allx â [a, b], by the Fundamental Theorem of Calculus.
Then, by the Additivity Theorem,
F (x) +H(x) =
â« x
a
f +
â« x
b
f =
â« b
a
f.
In particular,
H(x) =
â« b
a
f â F (x).
As F is differentiable,â« baf â F (x) is also differentiable; so is H since
H âČ(x) = 0â F âČ(x) = âf(x). ïżœ
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20. Proof. As f is continuous, F (x) =â« x0f is continuous and F âČ(x) = f(x)
for all x â [0,â) by the Fundamental Theorem of Calculus (2nd version),
Either f(x) > 0 for all x > 0 or f(x) < 0 for all x > 0 â otherwise fwould admit a root c > 0 by the IVT, which contradicts f(x) 6= 0 âx > 0.
But
F (x) =
â« x
0
f =(f(x))2
2> 0 for all x > 0,
soâ« x0f > 0 for all x > 0, i.e. f > 0 for all x > 0 â otherwise,â« x
0f â€
â« x00 = 0, which contradicts one of the above inequalities.
By construction,(f(0))2
2= F (0) =
â« 0
0
f = 0,
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that is f(0) = 0. Now, let c > 0. By hypothesis, F âČ(c) = f(c) > 0.
Furthermore, F (c) = (f(c))2
2 . As f is continuous at c,
limxâc
1
2(f(x) + f(c)) = f(c).
Thus we have:
1 =F âČ(c)
f(c)=
limxâc
F (x)â F (c)xâ c
limxâc
1
2(f(x) + f(c))
= limxâc
(f(x))2 â (f(c))
2
(xâ c) (f(x) + f(c))
= limxâc
(f(x)â f(c)) (f(x) + f(c))
(xâ c) (f(x) + f(c))= limxâc
f(x)â f(c)xâ c
= f âČ(c).
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Hence f is differentiable and f âČ(c) = 1 for all c > 0. Then, by theFundamental Theorem of Calculus (1st version),â« x
0
f âČ = f(x)â f(0) = f(x)â 0 = f(x)
for all x â„ 0. Asâ« x0f âČ =
â« x01 = x â 0 = x, this completes the proof,
which, incidentally, is one of my favourite proof in the course. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
21. Proof. As f and g are continuous, the functions
F (x) =
â« x
a
f and G(x) =
â« x
a
g
are continuous and differentiable on [a, b], with F âČ(x) = f(x) andGâČ(x) = g(x), according to the Fundamental Theorem of Calculus.Then H(x) = F (x)âG(x) is continuous.
But by hypothesis,
H(a) = F (a)âG(a) =â« a
a
f ââ« a
a
g = 0â 0 = 0
H(b) = F (b)âG(b) =â« b
a
f ââ« b
a
g = 0.
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
Since H is also differentiable, âc â (a, b) s.t. H âČ(c) = 0, by RolleâsTheorem. As
H âČ(c) = F âČ(c)âGâČ(c) = f(c)â g(c) = 0,
this completes the proof. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
22. Proof. As f is continuous, then F (x) =â« x0f is continuous and
differentiable on [0, 1], with F âČ(x) = f(x), by the Fundamental Theoremof Calculus.
By the Additivity Theorem,â« x
0
f +
â« 1
x
f =
â« 1
0
f.
Butâ« x0f =
â« 1
xf so 2
â« x0f =
â« 1
0f. In particular,
F (x) =1
2
â« 1
0
f = constant.
Then f(x) = F âČ(x) = 0 for all x â [0, 1]. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
23. Proof. The function f is increasing on [0, 3] so it is Riemann integrablethere. The function F is given by
F (x) =
x2
2 , x â [0, 1)
xâ 12, x â [1, 2)
x2â12 , x â [2, 3]
By the Fundamental Theorem of Calculus, F is differentiable whereverf is continuous, that is on [0, 2) âȘ (2, 3], and F âČ = f there. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
24. Proof. We will show the contrapositive of the statement.
Suppose that there exists z â [a, b] such that f(z) > 0. Since f iscontinuous, we may assume z â (a, b), as if f(z) = 0 for all z â (a, b),then f(a) = f(b) = 0.
Then, taking Δ = f(z)/2 in the definition of continuity, there existsa Ύ > 0 such that
|xâ z| < ÎŽ =â |f(x)â f(z)| < f(z)/2 =â f(x) > f(z)/2.
Reducing ÎŽ if necessary, we may assume ÎŽ †min{zâa, bâa}. Therefore,
[z â ÎŽ/2, z + ÎŽ/2] â (z â ÎŽ, z + ÎŽ) â [a, b].
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Thusâ« b
a
f =
â« zâÎŽ/2
a
f +
â« z+ÎŽ/2
zâÎŽ/2f +
â« b
z+ÎŽ/2
f â„ 0 + ÎŽf(z)/2 + 0 > 0.
This completes the proof. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
25. Proof. We will show the contrapositive of the statement.
Suppose f(c) 6= 0 for all c â [a, b]. Then, by the Intermediate ValueTheorem, either f(x) > 0 for all x â [a, b] or f(x) < 0 for all x â [a, b].
If f(x) > 0 for all x â [a, b], thenâ« baf > 0 by the preceding question.
Similarly, if f(x) < 0 for all x â [a, b], thenâ« ba(âf) > 0, which implies
that ââ« baf > 0. In both cases,
â« baf 6= 0. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
26. Solution. According to the additivity property of the Riemann integraland the Fundamental Theorem of Calculus, we have
d
dx
â« x
âxet
2dt =
d
dx
(â« 0
âxet
2dt+
â« x
0
et2dt
)=
d
dx
(ââ« âx0
et2dt+
â« x
0
et2dt
)= â d
dx
â« âx0
et2dt+
d
dx
â« x
0
et2dt
= âex2· (â1) + ex
2= 2ex
2,
where we used the chain rule in the second-to-last equality. ïżœ
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Solution. By definition,
â« 1
0
1â|x|
dx = limaâ0+
â« 1
a
1âxdx = lim
aâ0+
(2â1â 2
âa)= 2.
Thus the improper integral converges to 2. ïżœ
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MAT 2125 â Elementary Real Analysis Chapter 5 â Differential and Integral Calculus
27. Solution. If s â„ 0, the integral is not improper. The integrand beingcontinuous, the integral exists (i.e. converges) for those values of s.
Now, consider s < 0. First assume s 6= â1. We haveâ« 1
0
xs dx = limaâ0+
â« 1
a
xs dx = limaâ0+
(1s+1
s+ 1â as+1
s+ 1
).
This limit exists if and only if s > â1, in which case it is equal to 1s+1.
Now, if s = â1, then we haveâ« 1
0
xs dx = limaâ0+
â« 1
a
xâ1 dx = limaâ0+
(log 1â log a
),
which does not exist. Therefore, the given improper integral convergesif and only if s > â1. ïżœ
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