Lesson 25: Evaluating Definite Integrals (slides)

..

Sec on 5.3Evalua ng Definite Integrals

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

April 27, 2011

AnnouncementsI Today: 5.3I Thursday/Friday: Quiz on4.1–4.4

I Monday 5/2: 5.4I Wednesday 5/4: 5.5I Monday 5/9: Review andMovie Day!

I Thursday 5/12: FinalExam, 2:00–3:50pm

ObjectivesI Use the Evalua onTheorem to evaluatedefinite integrals.

I Write an deriva ves asindefinite integrals.

I Interpret definiteintegrals as “net change”of a func on over aninterval.

OutlineLast me: The Definite Integral

The definite integral as a limitProper es of the integral

Evalua ng Definite IntegralsExamples

The Integral as Net Change

Indefinite IntegralsMy first table of integrals

Compu ng Area with integrals

The definite integral as a limitDefini onIf f is a func on defined on [a, b], the definite integral of f from a tob is the number ∫ b

af(x) dx = lim

n→∞

n∑i=1

f(ci)∆x

where∆x =b− an

, and for each i, xi = a+ i∆x, and ci is a point in[xi−1, xi].

The definite integral as a limit

TheoremIf f is con nuous on [a, b] or if f has only finitely many jumpdiscon nui es, then f is integrable on [a, b]; that is, the definite

integral∫ b

af(x) dx exists and is the same for any choice of ci.

Notation/Terminology∫ b

af(x) dx

I

∫— integral sign (swoopy S)

I f(x)— integrandI a and b— limits of integra on (a is the lower limit and b theupper limit)

I dx— ??? (a parenthesis? an infinitesimal? a variable?)I The process of compu ng an integral is called integra on

Example

Es mate∫ 1

0

41+ x2

dx usingM4.

Solu on

Example

Es mate∫ 1

0

41+ x2

dx usingM4.

Solu on

We have x0 = 0, x1 =14, x2 =

12, x3 =

34, x4 = 1.

So c1 =18, c2 =

38, c3 =

58, c4 =

78.

Example

Es mate∫ 1

0

41+ x2

dx usingM4.

Solu on

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)

=14

(4

65/64+

473/64

+4

89/64+

4113/64

)=

6465

+6473

+6489

+64113

≈ 3.1468

Example

Es mate∫ 1

0

41+ x2

dx usingM4.

Solu on

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)=

14

(4

65/64+

473/64

+4

89/64+

4113/64

)

=6465

+6473

+6489

+64113

≈ 3.1468

Example

Es mate∫ 1

0

41+ x2

dx usingM4.

Solu on

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)=

14

(4

65/64+

473/64

+4

89/64+

4113/64

)=

6465

+6473

+6489

+64113

≈ 3.1468

Properties of the integralTheorem (Addi ve Proper es of the Integral)

Let f and g be integrable func ons on [a, b] and c a constant. Then

1.∫ b

ac dx = c(b− a)

2.∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx.

3.∫ b

acf(x) dx = c

∫ b

af(x) dx.

4.∫ b

a[f(x)− g(x)] dx =

∫ b

af(x) dx−

∫ b

ag(x) dx.

More Properties of the IntegralConven ons: ∫ a

bf(x) dx = −

∫ b

af(x) dx∫ a

af(x) dx = 0

This allows us to haveTheorem

5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c

bf(x) dx for all a, b, and c.

Illustrating Property 5Theorem

5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a

..b

..c


5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a

..b

..c

.

∫ b

af(x) dx


5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a

..b

..c

.

∫ b

af(x) dx

.

∫ c

bf(x) dx


5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a

..b

..c

.

∫ b

af(x) dx

.

∫ c

bf(x) dx

.

∫ c

af(x) dx


5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a..

b..

c


5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a..

b..

c.

∫ b

af(x) dx


5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a..

b..

c.

∫ c

bf(x) dx =

−∫ b

cf(x) dx


5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a..

b..

c.

∫ c

bf(x) dx =

−∫ b

cf(x) dx

.

∫ c

af(x) dx

Definite Integrals We Know So FarI If the integral computes an areaand we know the area, we canuse that. For instance,∫ 1

0

√1− x2 dx =

π

4

I By brute force we computed∫ 1

0x2 dx =

13

∫ 1

0x3 dx =

14

..x

.

y

Comparison Properties of the IntegralTheoremLet f and g be integrable func ons on [a, b].

6. If f(x) ≥ 0 for all x in [a, b], then∫ b

af(x) dx ≥ 0

7. If f(x) ≥ g(x) for all x in [a, b], then∫ b

af(x) dx ≥

∫ b

ag(x) dx

8. If m ≤ f(x) ≤ M for all x in [a, b], then

m(b− a) ≤∫ b

af(x) dx ≤ M(b− a)



af(x) dx ≥ 0


af(x) dx ≥

∫ b

ag(x) dx


m(b− a) ≤∫ b




af(x) dx ≥ 0


af(x) dx ≥

∫ b

ag(x) dx


m(b− a) ≤∫ b




af(x) dx ≥ 0


af(x) dx ≥

∫ b

ag(x) dx


m(b− a) ≤∫ b


Integral of a nonnegative function is nonnegativeProof.If f(x) ≥ 0 for all x in [a, b], then forany number of divisions n and choiceof sample points {ci}:

Sn =n∑

i=1

f(ci)︸︷︷︸≥0

∆x ≥n∑

i=1

0 ·∆x = 0

.. x.......Since Sn ≥ 0 for all n, the limit of {Sn} is nonnega ve, too:∫ b

af(x) dx = lim

n→∞Sn︸︷︷︸≥0

≥ 0

The integral is “increasing”Proof.Let h(x) = f(x)− g(x). If f(x) ≥ g(x)for all x in [a, b], then h(x) ≥ 0 for allx in [a, b]. So by the previousproperty ∫ b

ah(x) dx ≥ 0 .. x.

f(x)

.

g(x)

.

h(x)

This means that∫ b

af(x) dx−

∫ b

ag(x) dx =

∫ b

a(f(x)− g(x)) dx =

∫ b

ah(x) dx ≥ 0

Bounding the integralProof.Ifm ≤ f(x) ≤ M on for all x in [a, b], then bythe previous property∫ b

amdx ≤

∫ b

af(x) dx ≤

∫ b

aMdx

By Property 8, the integral of a constantfunc on is the product of the constant andthe width of the interval. So:

m(b− a) ≤∫ b


.. x.

y

.

M

.

f(x)

.

m

..a

..b

Example

Es mate∫ 2

1

1xdx using the comparison proper es.

Solu onSince

12≤ 1

x≤ 1

1for all x in [1, 2], we have

12· 1 ≤

∫ 2

1

1xdx ≤ 1 · 1

Example

Es mate∫ 2

1


Solu onSince

12≤ 1

x≤ 1


12· 1 ≤

∫ 2

1

1xdx ≤ 1 · 1

Ques on

Es mate∫ 2

1

1xdx with L2 and R2. Are your es mates overes mates?

Underes mates? Impossible to tell?

AnswerSince the integrand is decreasing,

Rn <

∫ 2

1

1xdx < Ln

for all n. So712

<

∫ 2

1

1xdx <

56.

Ques on

Es mate∫ 2

1

1xdx with L2 and R2. Are your es mates overes mates?

Underes mates? Impossible to tell?

AnswerSince the integrand is decreasing,

Rn <

∫ 2

1

1xdx < Ln

for all n. So712

<

∫ 2

1

1xdx <

56.







Socratic proofI The definite integral of velocitymeasures displacement (netdistance)

I The deriva ve of displacementis velocity

I So we can computedisplacement with the definiteintegral or the an deriva ve ofvelocity

I But any func on can be avelocity func on, so . . .

Theorem of the DayTheorem (The Second Fundamental Theorem of Calculus)

Suppose f is integrable on [a, b] and f = F′ for another func on F,then ∫ b

af(x) dx = F(b)− F(a).

NoteIn Sec on 5.3, this theorem is called “The Evalua on Theorem”.Nobody else in the world calls it that.

Theorem of the DayTheorem (The Second Fundamental Theorem of Calculus)

Suppose f is integrable on [a, b] and f = F′ for another func on F,then ∫ b

af(x) dx = F(b)− F(a).

NoteIn Sec on 5.3, this theorem is called “The Evalua on Theorem”.Nobody else in the world calls it that.

Proving the Second FTCProof.

I Divide up [a, b] into n pieces of equal width∆x =b− an

asusual.



asusual.

I For each i, F is con nuous on [xi−1, xi] and differen able on(xi−1, xi). So there is a point ci in (xi−1, xi) with

F(xi)− F(xi−1)

xi − xi−1= F′(ci) = f(ci)

= F(xi)− F(xi−1)



asusual.

I For each i, F is con nuous on [xi−1, xi] and differen able on(xi−1, xi). So there is a point ci in (xi−1, xi) with

F(xi)− F(xi−1)

xi − xi−1= F′(ci) = f(ci)

=⇒ f(ci)∆x = F(xi)− F(xi−1)


I Form the Riemann Sum:

Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))

= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))

= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)



Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


I We have shown for each n,

Sn = F(b)− F(a)

Which does not depend on n.

I So in the limit∫ b

af(x) dx = lim

n→∞Sn = lim

n→∞(F(b)− F(a)) = F(b)− F(a)


I We have shown for each n,

Sn = F(b)− F(a)

Which does not depend on n.I So in the limit∫ b

af(x) dx = lim

n→∞Sn = lim

n→∞(F(b)− F(a)) = F(b)− F(a)

Computing area with the 2nd FTCExample

Find the area between y = x3 and the x-axis, between x = 0 andx = 1.

Solu on

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10=

14

.

Here we use the nota on F(x)|ba or [F(x)]ba to mean F(b)− F(a).



Solu on

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10=

14 .




Solu on

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10=

14 .



Find the area enclosed by the parabola y = x2 and the line y = 1.

Solu on

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43 ...

−1..

1..

1



Solu on

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43

...−1

..1

..

1



Solu on

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43 ...

−1..

1..

1

Computing an integral weestimated before

Example

Evaluate the integral∫ 1

0

41+ x2

dx.

Solu on∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx

= 4 arctan(x)|10

= 4 (arctan 1− arctan 0) = 4(π4− 0

)

= π

Example

Es mate∫ 1

0

41+ x2

dx usingM4.

Solu on

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)=

14

(4

65/64+

473/64

+4

89/64+

4113/64

)=

6465

+6473

+6489

+64113

≈ 3.1468


Example


0

41+ x2

dx.

Solu on∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx

= 4 arctan(x)|10

= 4 (arctan 1− arctan 0) = 4(π4− 0

)

= π


Example


0

41+ x2

dx.

Solu on∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx = 4 arctan(x)|10

= 4 (arctan 1− arctan 0) = 4(π4− 0

)

= π


Example


0

41+ x2

dx.

Solu on∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx = 4 arctan(x)|10

= 4 (arctan 1− arctan 0)

= 4(π4− 0

)

= π


Example


0

41+ x2

dx.

Solu on∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx = 4 arctan(x)|10

= 4 (arctan 1− arctan 0) = 4(π4− 0

)

= π


Example


0

41+ x2

dx.

Solu on∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx = 4 arctan(x)|10

= 4 (arctan 1− arctan 0) = 4(π4− 0

)= π


Example

Evaluate∫ 2

1

1xdx.

Solu on ∫ 2

1

1xdx

= ln x|21 = ln 2− ln 1 = ln 2

Example

Es mate∫ 2

1


Solu onSince

12≤ 1

x≤ 1


12· 1 ≤

∫ 2

1

1xdx ≤ 1 · 1


Example

Evaluate∫ 2

1

1xdx.

Solu on ∫ 2

1

1xdx

= ln x|21 = ln 2− ln 1 = ln 2


Example

Evaluate∫ 2

1

1xdx.

Solu on ∫ 2

1

1xdx = ln x|21

= ln 2− ln 1 = ln 2


Example

Evaluate∫ 2

1

1xdx.

Solu on ∫ 2

1

1xdx = ln x|21 = ln 2− ln 1

= ln 2


Example

Evaluate∫ 2

1

1xdx.

Solu on ∫ 2

1

1xdx = ln x|21 = ln 2− ln 1 = ln 2








Another way to state this theorem is:∫ b

aF′(x) dx = F(b)− F(a),

or the integral of a deriva ve along an interval is the net changeover that interval. This has many interpreta ons.



Corollary

If v(t) represents the velocity of a par cle moving rec linearly, then∫ t1

t0v(t) dt = s(t1)− s(t0).


Corollary

If MC(x) represents the marginal cost of making x units of a product,then

C(x) = C(0) +∫ x

0MC(q) dq.


Corollary

If ρ(x) represents the density of a thin rod at a distance of x from itsend, then the mass of the rod up to x is

m(x) =∫ x

0ρ(s) ds.







A new notation for antiderivativesTo emphasize the rela onship between an differen a on andintegra on, we use the indefinite integral nota on∫

f(x) dx

for any func on whose deriva ve is f(x).

Thus∫x2 dx = 1

3x3 + C.

A new notation for antiderivativesTo emphasize the rela onship between an differen a on andintegra on, we use the indefinite integral nota on∫

f(x) dx

for any func on whose deriva ve is f(x). Thus∫x2 dx = 1

3x3 + C.

My first table of integrals..∫

[f(x) + g(x)] dx =∫

f(x) dx+∫

g(x) dx∫xn dx =

xn+1

n+ 1+ C (n ̸= −1)∫

ex dx = ex + C∫sin x dx = − cos x+ C∫cos x dx = sin x+ C∫sec2 x dx = tan x+ C∫

sec x tan x dx = sec x+ C∫1

1+ x2dx = arctan x+ C

∫cf(x) dx = c

∫f(x) dx∫

1xdx = ln |x|+ C∫

ax dx =ax

ln a+ C∫

csc2 x dx = − cot x+ C∫csc x cot x dx = − csc x+ C∫

1√1− x2

dx = arcsin x+ C







Computing Area with integrals

Example

Find the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex.

Solu onThe answer is ∫ 4

1ex dx = ex|41 = e4 − e.

Computing Area with integrals

Example

Find the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex.

Solu onThe answer is ∫ 4

1ex dx = ex|41 = e4 − e.

Computing Area with integralsExample

Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.

Solu on

I Instead compute the area as

π

2−∫ π/2

0sin y dy =

π

2−[− cos x]π/20 =

π

2−1

..x

.

y

..1

..

π/2



Solu on

I The answer is∫ 1

0arcsin x dx, but

we do not know an an deriva vefor arcsin.


π

2−∫ π/2

0sin y dy =

π

2−[− cos x]π/20 =

π

2−1

..x

.

y

..1

..

π/2



Solu on


π

2−∫ π/2

0sin y dy

=π

2−[− cos x]π/20 =

π

2−1

..x

.

y

..1

..

π/2



Solu on


π

2−∫ π/2

0sin y dy =

π

2−[− cos x]π/20

=π

2−1

..x

.

y

..1

..

π/2



Solu on


π

2−∫ π/2

0sin y dy =

π

2−[− cos x]π/20 =

π

2−1

..x

.

y

..1

..

π/2

Example

Find the area between the graph of y = (x− 1)(x− 2), the x-axis,and the ver cal lines x = 0 and x = 3.

Solu on

.. x.

y

..1

..2

..3

Example


Solu onNo ce the func ony = (x− 1)(x− 2) is posi ve on [0, 1)and (2, 3], and nega ve on (1, 2).

.. x.

y

..1

..2

..3

Example


Solu on

A =

∫ 1

0(x2 − 3x+ 2) dx

−∫ 2

1(x2 − 3x+ 2) dx

+

∫ 3

2(x2 − 3x+ 2) dx

.. x.

y

..1

..2

..3

Example


Solu on

A =

∫ 1

0(x− 1)(x− 2) dx

−∫ 2

1(x− 1)(x− 2) dx

+

∫ 3

2(x− 1)(x− 2) dx

.. x.

y

..1

..2

..3

Example


Solu on

A =[13x

3 − 32x

2 + 2x]10

−[13x

3 − 32x

2 + 2x]21

+[13x

3 − 32x

2 + 2x]32

=116

.. x.

y

..1

..2

..3

Interpretation of “negative area”in motion

There is an analog in rectlinear mo on:

I

∫ t1

t0v(t) dt is net distance traveled.

I

∫ t1

t0|v(t)| dt is total distance traveled.

What about the constant?I It seems we forgot about the+C when we say for instance∫ 1

0x3 dx =

x4

4

∣∣∣∣10=

14− 0 =

14

I But no ce[x4

4+ C

]10=

(14+ C

)− (0+ C) =

14+ C− C =

14

no ma er what C is.I So in an differen a on for definite integrals, the constant isimmaterial.

SummaryI The second Fundamental Theorem of Calculus:∫ b

af(x) dx = F(b)− F(a)

where F′ = f.I Definite integrals represent net change of a func on over aninterval.

I We write an deriva ves as indefinite integrals∫

f(x) dx

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Lesson 25: Evaluating Definite Integrals (slides)