Final Exam Review (Integration)

Review for Final ExamIntegration

Math 1a

January 13, 2008

Announcements

I Office hours on the website (click “Exams”)

I Email your TF, CA, or me with questions

I Final: Thursday 9:15am in Hall B

Outline

The Riemann IntegralEstimating the integralProperties of the integralComparison Properties of the Integral

The Fundamental Theorem of CalculusStatementDifferentiation of functions defined by integralsProperties of the area functionThe Second FTCExamplesTotal ChangeIndefinite Integrals

Integration by SubstitutionSubstitution for Indefinite IntegralsSubstitution for Definite Integrals

The Riemann IntegralLearning Objectives

I Compute the definite integral using a limit of Riemann sums

I Estimate the definite integral using a Riemann sum (e.g.,Midpoint Rule)

I Reason with the definite integral using its elementaryproperties.

The Area Problem

Given a function f defined on [a, b], how can one find the areabetween y = 0, y = f (x), x = a, and x = b?We divide and conquer.

Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f (x0)∆x + f (x1)∆x + · · ·+ f (xn−1)∆x

Rn = f (x1)∆x + f (x2)∆x + · · ·+ f (xn)∆x

Mn = f

(x0 + x1

2

)∆x + f

(x1 + x2

2

)∆x + · · ·+ f

(xn−1 + xn

2

)∆x

In general, choose ci to be a point in the ith interval [xi−1, xi ].Form the Riemann sum

Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x

=n∑

i=1

f (ci )∆x

Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f (x0)∆x + f (x1)∆x + · · ·+ f (xn−1)∆x

Rn = f (x1)∆x + f (x2)∆x + · · ·+ f (xn)∆x

Mn = f

(x0 + x1

2

)∆x + f

(x1 + x2

2

)∆x + · · ·+ f

(xn−1 + xn

2

)∆x

In general, choose ci to be a point in the ith interval [xi−1, xi ].Form the Riemann sum

Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x

=n∑

i=1

f (ci )∆x

Theorem

TheoremIf f is a continuous function on [a, b] or has finitely many jumpdiscontinuities, then

limn→∞

Sn = limn→∞

{f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x}

exists and is the same value no matter what choice of ci we made.

DefinitionThe definite integral of f from a to b is the number∫ b

af (x) dx = lim

∆x→0

n∑i=1

f (ci ) ∆x

Theorem

TheoremIf f is a continuous function on [a, b] or has finitely many jumpdiscontinuities, then

limn→∞

Sn = limn→∞

{f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x}

exists and is the same value no matter what choice of ci we made.

DefinitionThe definite integral of f from a to b is the number∫ b

af (x) dx = lim

∆x→0

n∑i=1

f (ci ) ∆x

Example (“Sample Exam”, Problem 6)

The rate at which the world’s oil is being consumed is increasing.Suppose that the rate (measured in billions of barrels per year) isgiven by the function r(t), where t is measured in years and t = 0represents January 1, 2000.

(a) Write a definite integral that represents the total quantity ofoil used between the start of 2000 and the start of 2005.

(b) Suppose that r(t) = 32e0.05t . Find the approximate value forthe definite integral from part (a) using a right-hand sum withn = 5 subintervals.

(c) Interpret each of the five terms in the sum from part (b) interms of oil consumption.

Answers

(a)

∫ 5

0r(t) dt

(b)

1·32e0.05(1)+1·32e0.05(2)+1·32e0.05(3)+1·32e0.05(4)+1·32e0.05(5)

(c) Each term stands for the approximate amount of oil used ineach year. For instance, the term 1 · 32e0.05(3) is approximatelythe amount of oil used between January 1, 2002 and January1, 2003.

Example

Estimate

∫ 1

0

4

1 + x2dx using the midpoint rule and four divisions.

Solution

The partition is 0 <1

4<

1

2<

3

4< 1, so the estimate is

M4 =1

4

(4

1 + (1/8)2+

4

1 + (3/8)2+

4

1 + (5/8)2+

4

1 + (7/8)2

)

=1

4

(4

65/64+

4

73/64+

4

89/64+

4

113/64

)=

150, 166, 784

47, 720, 465≈ 3.1468

Example

Estimate

∫ 1

0

4


Solution


4<

1

2<

3


M4 =1

4

(4

1 + (1/8)2+

4

1 + (3/8)2+

4

1 + (5/8)2+

4

1 + (7/8)2

)

=1

4

(4

65/64+

4

73/64+

4

89/64+

4

113/64

)=

150, 166, 784

47, 720, 465≈ 3.1468

Example

Estimate

∫ 1

0

4


Solution


4<

1

2<

3


M4 =1

4

(4

1 + (1/8)2+

4

1 + (3/8)2+

4

1 + (5/8)2+

4

1 + (7/8)2

)=

1

4

(4

65/64+

4

73/64+

4

89/64+

4

113/64

)

=150, 166, 784

47, 720, 465≈ 3.1468

Example

Estimate

∫ 1

0

4


Solution


4<

1

2<

3


M4 =1

4

(4

1 + (1/8)2+

4

1 + (3/8)2+

4

1 + (5/8)2+

4

1 + (7/8)2

)=

1

4

(4

65/64+

4

73/64+

4

89/64+

4

113/64

)=

150, 166, 784

47, 720, 465≈ 3.1468

Properties of the integral

Theorem (Additive Properties of the Integral)

Let f and g be integrable functions on [a, b] and c a constant.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.




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.




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.




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.

More Properties of the Integral

Conventions: ∫ a

bf (x) dx = −

∫ b

af (x) dx

∫ a

af (x) dx = 0

This allows us to have

5.

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx for all a, b, and c .


Conventions: ∫ a

bf (x) dx = −

∫ b

af (x) dx

∫ a

af (x) dx = 0


5.

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c



Conventions: ∫ a

bf (x) dx = −

∫ b

af (x) dx

∫ a

af (x) dx = 0


5.

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c


Example

Suppose f and g are functions with

I

∫ 4

0f (x) dx = 4

I

∫ 5

0f (x) dx = 7

I

∫ 5

0g(x) dx = 3.

Find

(a)

∫ 5

0[2f (x)− g(x)] dx

(b)

∫ 5

4f (x) dx .

SolutionWe have

(a) ∫ 5

0[2f (x)− g(x)] dx = 2

∫ 5

0f (x) dx −

∫ 5

0g(x) dx

= 2 · 7− 3 = 11

(b) ∫ 5

4f (x) dx =

∫ 5

0f (x) dx −

∫ 4

0f (x) dx

= 7− 4 = 3

SolutionWe have

(a) ∫ 5

0[2f (x)− g(x)] dx = 2

∫ 5

0f (x) dx −

∫ 5

0g(x) dx

= 2 · 7− 3 = 11

(b) ∫ 5

4f (x) dx =

∫ 5

0f (x) dx −

∫ 4

0f (x) dx

= 7− 4 = 3

Comparison Properties of the Integral

TheoremLet f and g be integrable functions on [a, b].

6. If f (x) ≥ 0 for all x in [a, b], then∫ b

af (x) dx ≥ 0

7. If f (x) ≥ g(x) for all x in [a, b], then∫ b

af (x) dx ≥

∫ b

ag(x) dx

8. If m ≤ f (x) ≤ M for all x in [a, b], then

m(b − a) ≤∫ b

af (x) dx ≤ M(b − a)




af (x) dx ≥ 0


af (x) dx ≥

∫ b

ag(x) dx


m(b − a) ≤∫ b





af (x) dx ≥ 0


af (x) dx ≥

∫ b

ag(x) dx


m(b − a) ≤∫ b





af (x) dx ≥ 0


af (x) dx ≥

∫ b

ag(x) dx


m(b − a) ≤∫ b


Example

Estimate

∫ 4

1

1√x + sin2 πx

dx using the comparison properties.

Outline




The Fundamental Theorem of CalculusLearning Objectives

I State and use both fundamental theorems of calculus

I Understand the relationship between integration andantidifferentiation

I Use FTC to compute derivatives of integrals with functions inthe limits

I Use FTC to compute areas or other accumulations

Theorem (The First Fundamental Theorem of Calculus)

Let f be an integrable function on [a, b] and define

g(x) =

∫ x

af (t) dt.

If f is continuous at x in (a, b), then g is differentiable at x and

g ′(x) = f (x).

Example (Spring 2000 Final, Problem 7c)

Finddy

dxif y =

∫ 100

x3+x

√p2 − p dp

Solution

Let A(u) =

∫ u

1

√p2 − p dp. By the Fundamental Theorem of

Calculus, A′(u) =√

u2 − u. We have

y ′ =d

dx

[∫ 100

x3+x

√p2 − p dp

]=

d

dx

[∫ 100

1

√p2 − p dp −

∫ x3+x

1

√p2 − p dp

]

=d

dx

[A(100)− A(x3 + x)

]= −A′(x3 + x) · (3x2 + 1)

= −(3x2 + 1)√

(x3 + x)2 − (x3 + x).

Example (Spring 2000 Final, Problem 7c)

Finddy

dxif y =

∫ 100

x3+x

√p2 − p dp

Solution

Let A(u) =

∫ u

1

√p2 − p dp. By the Fundamental Theorem of

Calculus, A′(u) =√

u2 − u. We have

y ′ =d

dx

[∫ 100

x3+x

√p2 − p dp

]=

d

dx

[∫ 100

1

√p2 − p dp −

∫ x3+x

1

√p2 − p dp

]

=d

dx

[A(100)− A(x3 + x)

]= −A′(x3 + x) · (3x2 + 1)

= −(3x2 + 1)√

(x3 + x)2 − (x3 + x).

Facts about g from f

Let f be the function whose graph is given below.Suppose the the position at time t seconds of a particle moving

along a coordinate axis is s(t) =

∫ t

0f (x) dx meters. Use the

graph to answer the following questions.

1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

What is the particle’s velocityat time t = 5?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

What is the particle’s velocityat time t = 5?

SolutionRecall that by the FTC wehave

s ′(t) = f (t).

So s ′(5) = f (5) = 2.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

Is the acceleration of the par-ticle at time t = 5 positive ornegative?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

Is the acceleration of the par-ticle at time t = 5 positive ornegative?

SolutionWe have s ′′(5) = f ′(5), whichlooks negative from thegraph.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

What is the particle’s positionat time t = 3?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

What is the particle’s positionat time t = 3?

SolutionSince on [0, 3], f (x) = x, wehave

s(3) =

∫ 3

0x dx =

9

2.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

At what time during the first 9seconds does s have its largestvalue?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


Solution




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionThe critical points of s arethe zeros of s ′ = f .




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionBy looking at the graph, wesee that f is positive fromt = 0 to t = 6, then negativefrom t = 6 to t = 9.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionTherefore s is increasing on[0, 6], then decreasing on[6, 9]. So its largest value isat t = 6.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

Approximately when is the ac-celeration zero?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

Approximately when is the ac-celeration zero?

Solutions ′′ = 0 when f ′ = 0, whichhappens at t = 4 and t = 7.5(approximately)




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

When is the particle movingtoward the origin? Away fromthe origin?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionThe particle is moving awayfrom the origin when s > 0and s ′ > 0.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionSince s(0) = 0 and s ′ > 0 on(0, 6), we know the particle ismoving away from the originthen.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionAfter t = 6, s ′ < 0, so theparticle is moving toward theorigin.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)

On which side (positive or neg-ative) of the origin does theparticle lie at time t = 9?




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionWe have s(9) =∫ 6

0f (x) dx +

∫ 9

6f (x) dx,

where the left integral ispositive and the right integralis negative.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionIn order to decide whethers(9) is positive or negative,we need to decide if the firstarea is more positive than thesecond area is negative.




∫ t



1 2 3 4 5 6 7 8 9

1

2

3

4

• (1,1)

• (2,2)

• (3,3)• (5,2)


SolutionThis appears to be the case,so s(9) is positive.

Theorem (The Second Fundamental Theorem of Calculus)

Suppose f is integrable on [a, b] and f = F ′ for another function f ,then ∫ b

af (x) dx = F (b)− F (a).

Examples

Find the following integrals:

I

∫ 1

0x2 dx ,

∫ 1

0x3 dx ,

∫ 2

1xn dx (n 6= −1),

∫ 2

1

1

xdx

I

∫ π

0sin θ dθ,

∫ 1

0ex dx

The Integral as Total Change

Another way to state this theorem is:∫ b

aF ′(x) dx = F (b)− F (a),

or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:



aF ′(x) dx = F (b)− F (a),


TheoremIf v(t) represents the velocity of a particle moving rectilinearly,then ∫ t1

t0

v(t) dt = s(t1)− s(t0).



aF ′(x) dx = F (b)− F (a),


TheoremIf MC (x) represents the marginal cost of making x units of aproduct, then

C (x) = C (0) +

∫ x

0MC (q) dq.



aF ′(x) dx = F (b)− F (a),


TheoremIf ρ(x) represents the density of a thin rod at a distance of x fromits end, then the mass of the rod up to x is

m(x) =

∫ x

0ρ(s) ds.

A new notation for antiderivatives

To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫

f (x) dx

for any function whose derivative is f (x).

Thus∫x2 dx = 1

3 x3 + C .

A new notation for antiderivatives

To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫

f (x) dx

for any function whose derivative is f (x). Thus∫x2 dx = 1

3 x3 + C .

My first table of integrals∫[f (x) + g(x)] dx =

∫f (x) dx +

∫g(x) dx∫

xn dx =xn+1

n + 1+ C (n 6= −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∫

1

xdx = 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




Integration by SubstitutionLearning Objectives

I Given an integral and a specific substitution, perform thatsubstitution

I Use the substitution method to evaluate definite andindefinite integrals

Theorem (The Substitution Rule)

If u = g(x) is a differentiable function whose range is an interval Iand f is continuous on I , then∫

f (g(x))g ′(x) dx =

∫f (u) du

or ∫f (u)

du

dxdx =

∫f (u) du

This is the “anti” version of the chain rule.

Example

Find

∫xex2

dx

SolutionLet u = x2. Then du = 2x dx and x dx = 1

2 du. So∫xex2

dx = 12

∫eu du

= 12 eu + C

= 12 ex2

+ C

Example

Find

∫xex2

dx

SolutionLet u = x2. Then du = 2x dx and x dx = 1

2 du. So∫xex2

dx = 12

∫eu du

= 12 eu + C

= 12 ex2

+ C

Theorem (The Substitution Rule for Definite Integrals)

If g ′ is continuous and f is continuous on the range of u = g(x),then ∫ b

af (g(x))g ′(x) dx =

∫ g(b)

g(a)f (u) du.

Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)

First compute the indefinite integral

∫cos2 x sin x dx and then

evaluate. Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −

∫u2 du

= −13 u3 + C = −1

3 cos3 x + C .

Therefore ∫ π

0cos2 x sin x dx = −1

3 cos3 x∣∣π0

= 23 .

Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)



evaluate.

Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −

∫u2 du

= −13 u3 + C = −1

3 cos3 x + C .

Therefore ∫ π


3 cos3 x∣∣π0

= 23 .

Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)



evaluate. Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −

∫u2 du

= −13 u3 + C = −1

3 cos3 x + C .

Therefore ∫ π


3 cos3 x∣∣π0

= 23 .

Solution (Fast Way)

Do both the substitution and the evaluation at the same time.

Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du

=

∫ 1

−1u2 du

= 13 u3∣∣1−1

=2

3.

Solution (Fast Way)

Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1.

So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du

=

∫ 1

−1u2 du

= 13 u3∣∣1−1

=2

3.

Solution (Fast Way)

Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du

=

∫ 1

−1u2 du

= 13 u3∣∣1−1

=2

3.

Final Exam Review (Integration)

Technology

f x1 x

f c2 x

f c1 x

f x0 x

f x2 x

f x dx meters

denite integral of f

f cn x n