72
. . . . . . Section 5.3 Evaluating Definite Integrals V63.0121.027, Calculus I December 3, 2009 Announcements I Final Exam is Friday, December 18, 2:00–3:50pm I Final is cumulative; topics will be represented roughly according to time spent on them . . Image credit: docman

Lesson 27: Evaluating Definite Integrals

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The second Fundamental Theorem of Calculus makes calculating definite integrals a problem of antidifferentiation!(this slideshow has the typos fixed)

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Page 1: Lesson 27: Evaluating Definite Integrals

. . . . . .

Section5.3EvaluatingDefiniteIntegrals

V63.0121.027, CalculusI

December3, 2009

Announcements

I FinalExamisFriday, December18, 2:00–3:50pmI Finaliscumulative; topicswillberepresentedroughlyaccordingtotimespentonthem

..Imagecredit: docman

Page 2: Lesson 27: Evaluating Definite Integrals

. . . . . .

Outline

Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral

EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral

EvaluatingDefiniteIntegralsExamples

TheIntegralasTotalChange

IndefiniteIntegralsMyfirsttableofintegrals

ComputingAreawithintegrals

Page 3: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 4: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 5: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 6: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 7: Lesson 27: Evaluating Definite Integrals

. . . . . .

DefiniteIntegralsWeKnowSoFar

I Iftheintegralcomputesanareaandweknowthearea, wecanusethat. Forinstance,∫ 1

0

√1− x2 dx =

π

2

I Bybruteforcewecomputed∫ 1

0x2 dx =

13

∫ 1

0x3 dx =

14

..x

.y

Page 8: Lesson 27: Evaluating Definite Integrals

. . . . . .

Outline

Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral

EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral

EvaluatingDefiniteIntegralsExamples

TheIntegralasTotalChange

IndefiniteIntegralsMyfirsttableofintegrals

ComputingAreawithintegrals

Page 9: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 10: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 11: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 12: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 13: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 14: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 15: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 16: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 17: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 18: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 19: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 20: Lesson 27: Evaluating Definite Integrals

. . . . . .

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)

Page 21: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 22: Lesson 27: Evaluating Definite Integrals

. . . . . .

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)

Page 23: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 24: Lesson 27: Evaluating Definite Integrals

. . . . . .

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)

Page 25: Lesson 27: Evaluating Definite Integrals

. . . . . .

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)

Page 26: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 27: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 28: Lesson 27: Evaluating Definite Integrals

. . . . . .

Outline

Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral

EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral

EvaluatingDefiniteIntegralsExamples

TheIntegralasTotalChange

IndefiniteIntegralsMyfirsttableofintegrals

ComputingAreawithintegrals

Page 29: Lesson 27: Evaluating Definite Integrals

. . . . . .

Socraticproof

I Thedefiniteintegralofvelocitymeasuresdisplacement(netdistance)

I Thederivativeofdisplacementisvelocity

I Sowecancomputedisplacementwiththedefiniteintegral or theantiderivativeofvelocity

I Butanyfunctioncanbeavelocityfunction, so. . .

Page 30: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 31: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 32: Lesson 27: Evaluating Definite Integrals

. . . . . .

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)

Page 33: Lesson 27: Evaluating Definite Integrals

. . . . . .

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!

Page 34: Lesson 27: Evaluating Definite Integrals

. . . . . .

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!

Page 35: Lesson 27: Evaluating Definite Integrals

. . . . . .

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)

Page 36: Lesson 27: Evaluating Definite Integrals

. . . . . .

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).

Page 37: Lesson 27: Evaluating Definite Integrals

. . . . . .

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).

Page 38: Lesson 27: Evaluating Definite Integrals

. . . . . .

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).

Page 39: Lesson 27: Evaluating Definite Integrals

. . . . . .

ExampleFindtheareaenclosedbytheparabola y = x2 and y = 1.

...−1

..1

..1

Solution

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43

Page 40: Lesson 27: Evaluating Definite Integrals

. . . . . .

ExampleFindtheareaenclosedbytheparabola y = x2 and y = 1.

...−1

..1

..1

Solution

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43

Page 41: Lesson 27: Evaluating Definite Integrals

. . . . . .

ExampleFindtheareaenclosedbytheparabola y = x2 and y = 1.

...−1

..1

..1

Solution

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43

Page 42: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 43: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 44: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 45: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 46: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 47: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 48: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 49: Lesson 27: Evaluating Definite Integrals

. . . . . .

Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx

= ln x|21

= ln 2− ln 1

= ln 2

Page 50: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 51: Lesson 27: Evaluating Definite Integrals

. . . . . .

Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx

= ln x|21

= ln 2− ln 1

= ln 2

Page 52: Lesson 27: Evaluating Definite Integrals

. . . . . .

Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx = ln x|21

= ln 2− ln 1

= ln 2

Page 53: Lesson 27: Evaluating Definite Integrals

. . . . . .

Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx = ln x|21

= ln 2− ln 1

= ln 2

Page 54: Lesson 27: Evaluating Definite Integrals

. . . . . .

Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx = ln x|21

= ln 2− ln 1

= ln 2

Page 55: Lesson 27: Evaluating Definite Integrals

. . . . . .

Outline

Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral

EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral

EvaluatingDefiniteIntegralsExamples

TheIntegralasTotalChange

IndefiniteIntegralsMyfirsttableofintegrals

ComputingAreawithintegrals

Page 56: Lesson 27: Evaluating Definite Integrals

. . . . . .

TheIntegralasTotalChange

Anotherwaytostatethistheoremis:∫ b

aF′(x)dx = F(b)− F(a),

or theintegralofaderivativealonganintervalisthetotalchangebetweenthesidesofthatinterval. Thishasmanyramifications:

Page 57: Lesson 27: Evaluating Definite Integrals

. . . . . .

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).

Page 58: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 59: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 60: Lesson 27: Evaluating Definite Integrals

. . . . . .

Outline

Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral

EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral

EvaluatingDefiniteIntegralsExamples

TheIntegralasTotalChange

IndefiniteIntegralsMyfirsttableofintegrals

ComputingAreawithintegrals

Page 61: Lesson 27: Evaluating Definite Integrals

. . . . . .

A newnotationforantiderivatives

Toemphasizetherelationshipbetweenantidifferentiationandintegration, weusethe indefiniteintegral notation∫

f(x)dx

foranyfunctionwhosederivativeis f(x).

Thus∫x2 dx = 1

3x3 + C.

Page 62: Lesson 27: Evaluating Definite Integrals

. . . . . .

A newnotationforantiderivatives

Toemphasizetherelationshipbetweenantidifferentiationandintegration, weusethe indefiniteintegral notation∫

f(x)dx

foranyfunctionwhosederivativeis f(x). Thus∫x2 dx = 1

3x3 + C.

Page 63: Lesson 27: Evaluating Definite Integrals

. . . . . .

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

Page 64: Lesson 27: Evaluating Definite Integrals

. . . . . .

Outline

Lasttime: TheDefiniteIntegralThedefiniteintegralasalimitPropertiesoftheintegral

EstimatingtheDefiniteIntegralTheMidpointRuleComparisonPropertiesoftheIntegral

EvaluatingDefiniteIntegralsExamples

TheIntegralasTotalChange

IndefiniteIntegralsMyfirsttableofintegrals

ComputingAreawithintegrals

Page 65: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 66: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 67: Lesson 27: Evaluating Definite Integrals

. . . . . .

Graph

. .x

.y

..1

..2

..3

Page 68: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 69: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 70: Lesson 27: Evaluating Definite Integrals

. . . . . .

Interpretationof“negativearea”inmotion

Thereisananaloginrectlinearmotion:

I∫ t1

t0v(t)dt is net distancetraveled.

I∫ t1

t0|v(t)|dt is total distancetraveled.

Page 71: Lesson 27: Evaluating Definite Integrals

. . . . . .

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.

Page 72: Lesson 27: Evaluating Definite Integrals

. . . . . .

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