. . . . . .

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

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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!