. . . . . .
Section5.3EvaluatingDefiniteIntegrals
V63.0121.034, CalculusI
December2, 2009
Announcements
I FinalExamisFriday, December18, 2:00–3:50pmI Finaliscumulative; topicswillberepresentedroughlyaccordingtotimespentonthem
..Imagecredit: docman
. . . . . .
Outline
Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral
EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral
EvaluatingDefiniteIntegralsExamples
TheIntegralasTotalChange
IndefiniteIntegralsMyfirsttableofintegrals
ComputingAreawithintegrals
. . . . . .
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].
TheoremIf f iscontinuouson [a,b] orif f hasonlyfinitelymanyjumpdiscontinuities, then f isintegrableon [a,b]; thatis, thedefinite
integral∫ b
af(x)dx existsandisthesameforanychoiceof ci.
. . . . . .
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.
. . . . . .
DefiniteIntegralsWeKnowSoFar
I Iftheintegralcomputesanareaandweknowthearea, wecanusethat. Forinstance,∫ 1
0
√1− x2 dx =
π
2
I Bybruteforcewecomputed∫ 1
0x2 dx =
13
∫ 1
0x3 dx =
14
..x
.y
. . . . . .
Outline
Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral
EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral
EvaluatingDefiniteIntegralsExamples
TheIntegralasTotalChange
IndefiniteIntegralsMyfirsttableofintegrals
ComputingAreawithintegrals
. . . . . .
TheMidpointRuleGivenapartitionof [a,b] into n pieces, let x̄i bethemidpointof[xi−1, xi]. Define
Mn =n∑
i=1
f(x̄i)∆x.
Example
Cmpute M2 for∫ 1
0x2 dx.
Solution
M2 =12·(14
)2
+12·(34
)2
=516 . .
.12
.
.
.y = x2
.x
.y
. . . . . .
TheMidpointRuleGivenapartitionof [a,b] into n pieces, let x̄i bethemidpointof[xi−1, xi]. Define
Mn =n∑
i=1
f(x̄i)∆x.
Example
Cmpute M2 for∫ 1
0x2 dx.
Solution
M2 =12·(14
)2
+12·(34
)2
=516 . .
.12
.
.
.y = x2
.x
.y
. . . . . .
Whyaremidpointsoftenbetter?
Compare L2, R2, and M2 for∫ 1
0x2 dx =
13:
L2 =12· (0)2 + 1
2·(12
)2
=18= 0.125
R2 =12·(12
)2
+12· (1)2 =
58= 0.625
M2 =12·(14
)2
+12·(34
)2
=516
= 0.3125
. .
.12
.
.
.
.
.
.
.y = x2
.x
.y
Where f ismonotone, oneof Ln and Rn willbetoomuch, andtheothertwolittle. But Mn allowsoverestimatesandunderestimatestocounteract.
. . . . . .
Whyaremidpointsoftenbetter?
Compare L2, R2, and M2 for∫ 1
0x2 dx =
13:
L2 =12· (0)2 + 1
2·(12
)2
=18= 0.125
R2 =12·(12
)2
+12· (1)2 =
58= 0.625
M2 =12·(14
)2
+12·(34
)2
=516
= 0.3125
. .
.12
.
.
.
.
.
.
.y = x2
.x
.y
Where f ismonotone, oneof Ln and Rn willbetoomuch, andtheothertwolittle. But Mn allowsoverestimatesandunderestimatestocounteract.
. . . . . .
Whyaremidpointsoftenbetter?
Compare L2, R2, and M2 for∫ 1
0x2 dx =
13:
L2 =12· (0)2 + 1
2·(12
)2
=18= 0.125
R2 =12·(12
)2
+12· (1)2 =
58= 0.625
M2 =12·(14
)2
+12·(34
)2
=516
= 0.3125
. .
.12
.
.
.
.
.
.
.y = x2
.x
.y
Where f ismonotone, oneof Ln and Rn willbetoomuch, andtheothertwolittle. But Mn allowsoverestimatesandunderestimatestocounteract.
. . . . . .
Whyaremidpointsoftenbetter?
Compare L2, R2, and M2 for∫ 1
0x2 dx =
13:
L2 =12· (0)2 + 1
2·(12
)2
=18= 0.125
R2 =12·(12
)2
+12· (1)2 =
58= 0.625
M2 =12·(14
)2
+12·(34
)2
=516
= 0.3125 . .
.12
.
.
.
.
.
.
.y = x2
.x
.y
Where f ismonotone, oneof Ln and Rn willbetoomuch, andtheothertwolittle. But Mn allowsoverestimatesandunderestimatestocounteract.
. . . . . .
Whyaremidpointsoftenbetter?
Compare L2, R2, and M2 for∫ 1
0x2 dx =
13:
L2 =12· (0)2 + 1
2·(12
)2
=18= 0.125
R2 =12·(12
)2
+12· (1)2 =
58= 0.625
M2 =12·(14
)2
+12·(34
)2
=516
= 0.3125 . .
.12
.
.
.
.
.
.
.y = x2
.x
.y
Where f ismonotone, oneof Ln and Rn willbetoomuch, andtheothertwolittle. But Mn allowsoverestimatesandunderestimatestocounteract.
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx usingthemidpointruleandfourdivisions.
SolutionDividingup [0, 1] into 4 piecesgives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
Sothemidpointrulegives
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
)=
150, 166,78447, 720, 465
≈ 3.1468
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx usingthemidpointruleandfourdivisions.
SolutionDividingup [0, 1] into 4 piecesgives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
Sothemidpointrulegives
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
)=
150, 166,78447, 720, 465
≈ 3.1468
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx usingthemidpointruleandfourdivisions.
SolutionDividingup [0, 1] into 4 piecesgives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
Sothemidpointrulegives
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
)
=150, 166,78447, 720, 465
≈ 3.1468
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx usingthemidpointruleandfourdivisions.
SolutionDividingup [0, 1] into 4 piecesgives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
Sothemidpointrulegives
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
)=
150, 166,78447, 720, 465
≈ 3.1468
. . . . . .
ComparisonPropertiesoftheIntegralTheoremLet 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)
. . . . . .
Theintegralofanonnegativefunctionisnonnegative
Proof.If f(x) ≥ 0 forall x in [a,b], thenforanynumberofdivisions nandchoiceofsamplepoints {ci}:
Sn =n∑
i=1
f(ci)︸︷︷︸≥0
∆x ≥n∑
i=1
0 ·∆x = 0
Since Sn ≥ 0 forall n, thelimitof {Sn} isnonnegative, too:∫ b
af(x)dx = lim
n→∞Sn︸︷︷︸≥0
≥ 0
. . . . . .
ComparisonPropertiesoftheIntegralTheoremLet 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)
. . . . . .
Thedefiniteintegralis“increasing”
Proof.Let h(x) = f(x)− g(x). If f(x) ≥ g(x) forall x in [a,b], thenh(x) ≥ 0 forall x in [a,b]. Sobythepreviousproperty∫ b
ah(x)dx ≥ 0
Thismeansthat∫ b
af(x)dx−
∫ b
ag(x)dx =
∫ b
a(f(x)− g(x))dx =
∫ b
ah(x)dx ≥ 0
So ∫ b
af(x)dx ≥
∫ b
ag(x)dx
. . . . . .
ComparisonPropertiesoftheIntegralTheoremLet 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)
. . . . . .
Boundingtheintegralusingboundsofthefunction
Proof.If m ≤ f(x) ≤ M onforall x in [a,b], thenbythepreviousproperty∫ b
amdx ≤
∫ b
af(x)dx ≤
∫ b
aMdx
ByProperty 1, theintegralofaconstantfunctionistheproductoftheconstantandthewidthoftheinterval. So:
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
. . . . . .
Example
Estimate∫ 2
1
1xdx usingProperty 8.
SolutionSince
1 ≤ x ≤ 2 =⇒ 12≤ 1
x≤ 1
1wehave
12· (2− 1) ≤
∫ 2
1
1xdx ≤ 1 · (2− 1)
or12≤
∫ 2
1
1xdx ≤ 1
. . . . . .
Example
Estimate∫ 2
1
1xdx usingProperty 8.
SolutionSince
1 ≤ x ≤ 2 =⇒ 12≤ 1
x≤ 1
1wehave
12· (2− 1) ≤
∫ 2
1
1xdx ≤ 1 · (2− 1)
or12≤
∫ 2
1
1xdx ≤ 1
. . . . . .
Outline
Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral
EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral
EvaluatingDefiniteIntegralsExamples
TheIntegralasTotalChange
IndefiniteIntegralsMyfirsttableofintegrals
ComputingAreawithintegrals
. . . . . .
Socraticproof
I Thedefiniteintegralofvelocitymeasuresdisplacement(netdistance)
I Thederivativeofdisplacementisvelocity
I Sowecancomputedisplacementwiththedefiniteintegral or theantiderivativeofvelocity
I Butanyfunctioncanbeavelocityfunction, so. . .
. . . . . .
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.
. . . . . .
ProvingtheSecondFTC
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−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
SeeifyoucanspottheinvocationoftheMeanValueTheorem!
. . . . . .
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−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
SeeifyoucanspottheinvocationoftheMeanValueTheorem!
. . . . . .
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.
...−1
..1
..1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43
. . . . . .
ExampleFindtheareaenclosedbytheparabola y = x2 and y = 1.
...−1
..1
..1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43
. . . . . .
ExampleFindtheareaenclosedbytheparabola y = x2 and y = 1.
...−1
..1
..1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43
. . . . . .
Example
Evaluatetheintegral∫ 1
0
41+ x2
dx.
Solution
∫ 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
Estimate∫ 1
0
41+ x2
dx usingthemidpointruleandfourdivisions.
SolutionDividingup [0, 1] into 4 piecesgives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
Sothemidpointrulegives
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
)=
150, 166,78447, 720, 465
≈ 3.1468
. . . . . .
Example
Evaluatetheintegral∫ 1
0
41+ x2
dx.
Solution
∫ 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
Evaluatetheintegral∫ 1
0
41+ x2
dx.
Solution
∫ 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
Evaluatetheintegral∫ 1
0
41+ x2
dx.
Solution
∫ 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
Evaluatetheintegral∫ 1
0
41+ x2
dx.
Solution
∫ 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
Evaluatetheintegral∫ 1
0
41+ x2
dx.
Solution
∫ 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.
Solution
∫ 2
1
1xdx
= ln x|21
= ln 2− ln 1
= ln 2
. . . . . .
Example
Estimate∫ 2
1
1xdx usingProperty 8.
SolutionSince
1 ≤ x ≤ 2 =⇒ 12≤ 1
x≤ 1
1wehave
12· (2− 1) ≤
∫ 2
1
1xdx ≤ 1 · (2− 1)
or12≤
∫ 2
1
1xdx ≤ 1
. . . . . .
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx
= ln x|21
= ln 2− ln 1
= ln 2
. . . . . .
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx = ln x|21
= ln 2− ln 1
= ln 2
. . . . . .
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx = ln x|21
= ln 2− ln 1
= ln 2
. . . . . .
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx = ln x|21
= ln 2− ln 1
= ln 2
. . . . . .
Outline
Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral
EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral
EvaluatingDefiniteIntegralsExamples
TheIntegralasTotalChange
IndefiniteIntegralsMyfirsttableofintegrals
ComputingAreawithintegrals
. . . . . .
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: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral
EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral
EvaluatingDefiniteIntegralsExamples
TheIntegralasTotalChange
IndefiniteIntegralsMyfirsttableofintegrals
ComputingAreawithintegrals
. . . . . .
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: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral
EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral
EvaluatingDefiniteIntegralsExamples
TheIntegralasTotalChange
IndefiniteIntegralsMyfirsttableofintegrals
ComputingAreawithintegrals
. . . . . .
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.
. . . . . .
Graph
. .x
.y
..1
..2
..3
. . . . . .
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.
. . . . . .
Interpretationof“negativearea”inmotion
Thereisananaloginrectlinearmotion:
I∫ t1
t0v(t)dt is net distancetraveled.
I∫ t1
t0|v(t)|dt is total distancetraveled.
. . . . . .
Whatabouttheconstant?
I Itseemsweforgotaboutthe +C whenwesayforinstance∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14− 0 =
14
I Butnotice[x4
4+ C
]10=
(14+ C
)− (0+ C) =
14+ C− C =
14
nomatterwhat C is.I Soinantidifferentiation fordefiniteintegrals, theconstantisimmaterial.
. . . . . .
Whathavewelearnedtoday?
I Thesecond FundamentalTheoremofCalculus:∫ b
af(x)dx = F(b)− F(a)
where F′ = f.I Definiteintegralsrepresent netchange ofafunctionoveraninterval.
I Wewriteantiderivativesas indefiniteintegrals∫
f(x)dx