Lesson 25: Evaluating Definite Integrals (Section 10 version)
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Transcript of Lesson 25: Evaluating Definite Integrals (Section 10 version)
. . . . . .
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