lesson 27: evaluating definite integrals
DESCRIPTION
Also known as the second fundamental theorem of calculusTRANSCRIPT
. . . . . .
Section5.3EvaluatingDefiniteIntegrals
Math1aIntroductiontoCalculus
April16, 2008
Announcements
◮ Midtermisfinished: x̄ ≈ 43, σ ≈ 6.◮ MidtermIII isWednesday4/30inclass◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323
..Image: Flickruserdocman
. . . . . .
Announcements
◮ Midtermisfinished: x̄ ≈ 43, σ ≈ 6.◮ MidtermIII isWednesday4/30inclass◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323
. . . . . .
Outline
Lasttime: TheDefiniteIntegral
EvaluatingDefiniteIntegralsExamples
TotalChange
IndefiniteIntegralsMyfirsttableofintegrals
Examples“NegativeArea”
Midtermcomments
. . . . . .
Thedefiniteintegralasalimit
DefinitionIf f isafunctiondefinedon [a,b], the definiteintegralof f from ato b isthenumber∫ b
af(x)dx = lim
∆x→0
n∑i=1
f(ci) ∆x
. . . . . .
Notation/Terminology
∫ b
af(x)dx
◮∫
— integralsign (swoopy S)
◮ f(x) — integrand◮ a and b — limitsofintegration (a isthe lowerlimit and bthe upperlimit)
◮ dx —??? (aparenthesis? aninfinitesimal? avariable?)◮ 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”
Midtermcomments
. . . . . .
Socraticproof
◮ Thedefiniteintegralofvelocitymeasuresdisplacement(netdistance)
◮ Thederivativeofdisplacementisvelocity
◮ Sowecancomputedisplacementwiththeantiderivativeofvelocity?
. . . . . .
Theorem(TheSecondFundamentalTheoremofCalculus)Suppose f isintegrableon [a,b] and f = F′ foranotherfunction F,then ∫ b
af(x)dx = F(b) − F(a).
. . . . . .
Proof.WewillchooseRiemannsumswhichconvergetotheright-handside. Let n begiven. Ontheinterval [xi−1, xi] thereisapoint cisuchthat
f(ci) = F′(ci) =F(xi) − F(xi−1)
xi − xi−1.
Then
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)−F(xi−1)
)= F(xn)−F(x0) = F(b)−F(a).
So Sn → F(b) − F(a) as n → ∞.
. . . . . .
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
. . . . . .
Outline
Lasttime: TheDefiniteIntegral
EvaluatingDefiniteIntegralsExamples
TotalChange
IndefiniteIntegralsMyfirsttableofintegrals
Examples“NegativeArea”
Midtermcomments
. . . . . .
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”
Midtermcomments
. . . . . .
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”
Midtermcomments
. . . . . .
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
.
. . . . . .
Outline
Lasttime: TheDefiniteIntegral
EvaluatingDefiniteIntegralsExamples
TotalChange
IndefiniteIntegralsMyfirsttableofintegrals
Examples“NegativeArea”
Midtermcomments
. . . . . .
Midtermcomments
◮ x̄ ≈ 43, σ ≈ 6. So43-49isgood, 49+isgreat◮ I willpostsolutions. Pleasestudythem!◮ Remember“curve”isappliedattheend◮ Redemptionpolicy: yourfinalcancanceloutoneofyourmidterms!