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Computing integrals with Riemann sums is like computing derivatives with limits. The calculus of integrals turns out to come from antidifferentiation. This startling fact is the Second Fundamental Theorem of Calculus!
Citation preview
. . . . . .
Section5.3EvaluatingDefiniteIntegrals
V63.0121, CalculusI
April20, 2009
Announcements
I FinalExamisFriday, May8, 2:00–3:50pmI Finaliscumulative; topicswillberepresentedroughlyaccordingtotimespentonthem
..Imagecredit: docman
. . . . . .
Outline
Lasttime: TheDefiniteIntegral
EvaluatingDefiniteIntegralsExamples
TotalChange
IndefiniteIntegralsMyfirsttableofintegrals
Examples“NegativeArea”
. . . . . .
Thedefiniteintegralasalimit
DefinitionIf f isafunctiondefinedon [a,b], the definiteintegralof f from ato b isthenumber∫ b
af(x)dx = lim
n→∞
n∑i=1
f(ci)∆x
where ∆x =b− an
, andforeach i, xi = a + i∆x, and ci isapoint
in [xi−1, xi].
. . . . . .
Notation/Terminology
∫ b
af(x)dx
I∫
— integralsign (swoopy S)
I f(x) — integrandI a and b — limitsofintegration (a isthe lowerlimit and bthe upperlimit)
I dx —??? (aparenthesis? aninfinitesimal? avariable?)I Theprocessofcomputinganintegraliscalled integration
. . . . . .
Propertiesoftheintegral
Theorem(AdditivePropertiesoftheIntegral)Let f and g beintegrablefunctionson [a,b] and c aconstant.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.
. . . . . .
MorePropertiesoftheIntegral
Conventions: ∫ a
bf(x)dx = −
∫ b
af(x)dx∫ a
af(x)dx = 0
Thisallowsustohave
5.∫ c
af(x)dx =
∫ b
af(x)dx +
∫ c
bf(x)dx forall a, b, and c.
. . . . . .
ComparisonPropertiesoftheIntegral
TheoremLet f and g beintegrablefunctionson [a,b].
6. If f(x) ≥ 0 forall x in [a,b], then∫ b
af(x)dx ≥ 0
7. If f(x) ≥ g(x) forall x in [a,b], then∫ b
af(x)dx ≥
∫ b
ag(x)dx
8. If m ≤ f(x) ≤ M forall x in [a,b], then
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
. . . . . .
Outline
Lasttime: TheDefiniteIntegral
EvaluatingDefiniteIntegralsExamples
TotalChange
IndefiniteIntegralsMyfirsttableofintegrals
Examples“NegativeArea”
. . . . . .
Socraticproof
I Thedefiniteintegralofvelocitymeasuresdisplacement(netdistance)
I Thederivativeofdisplacementisvelocity
I Sowecancomputedisplacementwiththeantiderivativeofvelocity?
. . . . . .
TheoremoftheDay
Theorem(TheSecondFundamentalTheoremofCalculus)Suppose f isintegrableon [a,b] and f = F′ foranotherfunction F,then ∫ b
af(x)dx = F(b) − F(a).
NoteInSection5.3., thistheoremiscalled“TheEvaluationTheorem”.Nobodyelseintheworldcallsitthat.
. . . . . .
TheoremoftheDay
Theorem(TheSecondFundamentalTheoremofCalculus)Suppose f isintegrableon [a,b] and f = F′ foranotherfunction F,then ∫ b
af(x)dx = F(b) − F(a).
NoteInSection5.3., thistheoremiscalled“TheEvaluationTheorem”.Nobodyelseintheworldcallsitthat.
. . . . . .
Proving2FTC
Divideup [a,b] into n piecesofequalwidth ∆x =b− an
as
usual. Foreach i, F iscontinuouson [xi−1, xi] anddifferentiableon (xi−1, xi). Sothereisapoint ci in (xi−1, xi) with
F(xi) − F(xi−1)
xi − xi−1= F′(ci) = f(ci)
Orf(ci)∆x = F(xi) − F(xi−1)
. . . . . .
Wehaveforeach i
f(ci)∆x = F(xi) − F(xi−1)
FormtheRiemannSum:
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−1)) + (F(xn) − F(xn−1))
= F(xn) − F(x0) = F(b) − F(a)
. . . . . .
Wehaveshownforeach n,
Sn = F(b) − F(a)
sointhelimit∫ b
af(x)dx = lim
n→∞Sn = lim
n→∞(F(b) − F(a)) = F(b) − F(a)
. . . . . .
ExampleFindtheareabetween y = x3 andthe x-axis, between x = 0 andx = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10
=14
.
Hereweusethenotation F(x)|ba or [F(x)]ba tomean F(b) − F(a).
. . . . . .
ExampleFindtheareabetween y = x3 andthe x-axis, between x = 0 andx = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10
=14 .
Hereweusethenotation F(x)|ba or [F(x)]ba tomean F(b) − F(a).
. . . . . .
ExampleFindtheareabetween y = x3 andthe x-axis, between x = 0 andx = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10
=14 .
Hereweusethenotation F(x)|ba or [F(x)]ba tomean F(b) − F(a).
. . . . . .
ExampleFindtheareaenclosedbytheparabola y = x2 and y = 1.
.
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−−1
3
)]=
43
. . . . . .
ExampleFindtheareaenclosedbytheparabola y = x2 and y = 1.
.
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−−1
3
)]=
43
. . . . . .
ExampleFindtheareaenclosedbytheparabola y = x2 and y = 1.
.
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−−1
3
)]=
43
. . . . . .
Outline
Lasttime: TheDefiniteIntegral
EvaluatingDefiniteIntegralsExamples
TotalChange
IndefiniteIntegralsMyfirsttableofintegrals
Examples“NegativeArea”
. . . . . .
TheIntegralasTotalChange
Anotherwaytostatethistheoremis:∫ b
aF′(x)dx = F(b) − F(a),
or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:
. . . . . .
TheIntegralasTotalChange
Anotherwaytostatethistheoremis:∫ b
aF′(x)dx = F(b) − F(a),
or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:
TheoremIf v(t) representsthevelocityofaparticlemovingrectilinearly,then ∫ t1
t0v(t)dt = s(t1) − s(t0).
. . . . . .
TheIntegralasTotalChange
Anotherwaytostatethistheoremis:∫ b
aF′(x)dx = F(b) − F(a),
or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:
TheoremIf MC(x) representsthemarginalcostofmaking x unitsofaproduct, then
C(x) = C(0) +
∫ x
0MC(q)dq.
. . . . . .
TheIntegralasTotalChange
Anotherwaytostatethistheoremis:∫ b
aF′(x)dx = F(b) − F(a),
or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:
TheoremIf ρ(x) representsthedensityofathinrodatadistanceof x fromitsend, thenthemassoftherodupto x is
m(x) =
∫ x
0ρ(s)ds.
. . . . . .
Outline
Lasttime: TheDefiniteIntegral
EvaluatingDefiniteIntegralsExamples
TotalChange
IndefiniteIntegralsMyfirsttableofintegrals
Examples“NegativeArea”
. . . . . .
A newnotationforantiderivatives
Toemphasizetherelationshipbetweenantidifferentiationandintegration, weusethe indefiniteintegral notation∫
f(x)dx
foranyfunctionwhosederivativeis f(x).
Thus∫x2 dx = 1
3x3 + C.
. . . . . .
A newnotationforantiderivatives
Toemphasizetherelationshipbetweenantidifferentiationandintegration, weusethe indefiniteintegral notation∫
f(x)dx
foranyfunctionwhosederivativeis f(x). Thus∫x2 dx = 1
3x3 + C.
. . . . . .
Myfirsttableofintegrals∫[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
. . . . . .
Outline
Lasttime: TheDefiniteIntegral
EvaluatingDefiniteIntegralsExamples
TotalChange
IndefiniteIntegralsMyfirsttableofintegrals
Examples“NegativeArea”
. . . . . .
ExampleFindtheareabetweenthegraphof y = (x− 1)(x− 2), the x-axis,andtheverticallines x = 0 and x = 3.
Solution
Consider∫ 3
0(x− 1)(x− 2)dx. Noticetheintegrandispositiveon
[0, 1) and (2, 3], andnegativeon (1, 2). Ifwewanttheareaoftheregion, wehavetodo
A =
∫ 1
0(x− 1)(x− 2)dx−
∫ 2
1(x− 1)(x− 2)dx +
∫ 3
2(x− 1)(x− 2)dx
=[13x
3 − 32x
2 + 2x]10 −
[13x
3 − 32x
2 + 2x]21 +
[13x
3 − 32x
2 + 2x]32
=56−
(−16
)+
56
=116
.
. . . . . .
ExampleFindtheareabetweenthegraphof y = (x− 1)(x− 2), the x-axis,andtheverticallines x = 0 and x = 3.
Solution
Consider∫ 3
0(x− 1)(x− 2)dx. Noticetheintegrandispositiveon
[0, 1) and (2, 3], andnegativeon (1, 2). Ifwewanttheareaoftheregion, wehavetodo
A =
∫ 1
0(x− 1)(x− 2)dx−
∫ 2
1(x− 1)(x− 2)dx +
∫ 3
2(x− 1)(x− 2)dx
=[13x
3 − 32x
2 + 2x]10 −
[13x
3 − 32x
2 + 2x]21 +
[13x
3 − 32x
2 + 2x]32
=56−
(−16
)+
56
=116
.
. . . . . .
Graphfrompreviousexample
. .x
.y
..1
..2
..3
. . . . . .
Summary
I integralscanbecomputedwithantidifferentiationI integralofinstantaneousrateofchangeistotalnetchangeI ThesecondFunamentalTheoremofCalculusrequirestheMeanValueTheorem