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Test 3 Review Jiwen He Department of Mathematics, University of Houston [email protected] math.uh.edu/jiwenhe/Math1431 Jiwen He, University of Houston Math 1431 – Section 24076, Test 3 Review November 25, 2008 1 / 47

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Test 3 Review

Jiwen He

Department of Mathematics, University of Houston

[email protected]/∼jiwenhe/Math1431

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Test 3

Test 3: Dec. 4-6 in CASA

Material - Through 6.3.

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No Homework (Thanksgiving)

No homework this week!

Have a GREAT Thanksgiving!

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Final Exam

Final Exam: Dec. 14-17 in CASA

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You Might Be Interested to Know ...

I will replace your lowest test score with the percentage gradefrom the final exam (provided it is higher).

I will give an A to anyone who receives 95% or above on thefinal exam.

I will give a passing grade to anyone who receives at least70% on the final exam.

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Quiz 1

What is today?

a. Monday

b. Wednesday

c. Friday

d. None of these

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Test 3 Material

Test 3 will cover material from Chapter 5, along with Sections6.1, 6.2 and 6.3.

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Good Sources of Practice Problems

Examples from class.

The basic homework problems.

The basic online quiz problems.

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Definite Integral and Lower/Upper Sums

Area of Ω = Area of Ω1 + Area of Ω2 + · · ·+ Area of Ωn,

Lf (P) = m1∆x1 + m2∆x2 + · · ·+ mn∆xn

Uf (P) = M1∆x1 + M2∆x2 + · · ·+ Mn∆xn

Lf (P) ≤∫ b

af (x) dx ≤ Uf (P), for all partitions P of [a, b]

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Problem 1

Give both the upper and lower Riemann sums for the functionf (x) = x2 over the interval [1, 3] with respect to the partitionP = 1, 3

2 , 2, 3.

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Lower/Upper Sums and Riemann Sums

S∗(P) = f (x∗1 )∆x1 + f (x∗2 )∆x2 + · · ·+ f (x∗n )∆xn

Lf (P) ≤ S∗(P) ≤ Uf (P), for all partitions P of [a, b]

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Problem 2

Give the Riemann sums for the function f (x) = x2 over the interval[1, 3] with respect to the partition P = 1, 3

2 , 2, 3 using midpoints.

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Fundamental Theorem of Integral Calculus

Theorem

In general, ∫ b

af (x) dx = F (b)− F (a).

where F (x) is an antiderivative of f (x).

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Problem 3

Evaluate

1.

∫ 1

0

(2x − 6x4 + 5

)dx

2.

∫ 2

1

x4 + 1

x2dx

3.

∫ 1

0

(4−

√x)2

dx

4.

∫ π/4

0sec x

(2 tan x − 5 sec x

)dx

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Area below the graph of a Nonnegative f

f (x) ≥ 0 for all x in [a, b].

Ω = region below the graph of f .

Area of Ω =

∫ b

af (x) dx = F (b)− F (a)

where F (x) is an antiderivative of f (x).

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Problem 4

Find the area bounded above by the graph of f (x) = x2 and belowby the x-axis over the interval [1, 3].

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Area between the graphs of f and g

f (x) ≥ g(x) for all x in [a, b].

Ω = region between the graphs of f (Top) and g (Bottom).

Area of Ω =

∫ b

a

[Top − Bottom

]dx =

∫ b

a

[f (x)− g(x)

]dx .

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Problem 5

Find the area between the graphs of y = 4x and y = x3 over theinterval [−2, 2].

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∫ b

a f (x) dx as Signed Area

∫ b

af (x) dx =

∫ c

af (x) dx +

∫ d

cf (x) dx +

∫ e

df (x) dx +

∫ b

ef (x) dx

= Area of Ω1 − Area of Ω2 + Area of Ω3 − Area of Ω4

=[Area of Ω1 + Area of Ω3

]−[Area of Ω2 + Area of Ω4

]= Area above the x-axis− Area below the x-axis.

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Problem 6

The graph of y = f (x) is shown below. The region Ω2 has area 3

and

∫ c

af (x)dx is 2. Give the area of region Ω1.

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Problem 7

Give the area bounded between the graph of f (x) = x2 − 2xand the x-axis on [−1, 3].

Evaluate

∫ 3

−1

(x2 − 2x

)dx and interpret the result in terms of

areas.

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Indefinite Integral as General Antiderivative

The Indefinite Integral of f

In general, ∫f (x) dx = F (x) + C

where F (x) is any antiderivative of f (x) and C is an arbitraryconstant.

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Problem 8

1. Find F given that F ′(x) = cos 3x and F (−π) = 1.

2. Give an antiderivative of f (x) = cos 3x whose graph hasy -intercept 3.

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Undoing the Chain Rule: The u-Substitution

Note: Most differentiation involves the chain rule, so we shouldexpect that most antidifferentiation will involve

undoing the chain rule

(the u-Substitution)

If F is an antiderivative for f , then

d

dx

[F (u(x))

]= F ′(u(x)) u′(x) = f (u(x)) u′(x)∫

f (u(x)) u′(x) dx =

∫f (u) du = F (u) + C = F (u(x)) + C .

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The u-Substitution

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Problem 9

Calculate

1.

∫sin x cos x dx

2.

∫2x3 sec2(x4 + 1) dx

3.

∫sec3 x tan x dx

4.

∫x(x − 3)5 dx

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Substitution in Definite Integrals

The Change of Variables Formula∫ b

af (u(x))u′(x) dx =

∫ u(b)

u(a)f (u) du.

We change the limits of integration to reflect the substitution.

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Problem 10

Evaluate

1.

∫ 2

0(x2 − 1)(x3 − 3x + 2)3 dx

2.

∫ 1/2

0cos3 πx sin πx dx

3.

∫ √3

0x5√

x2 + 1 dx

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Definite Integral and Antiderivative

d

dx

(∫ x

af (t) dt

)= f (x).

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Problem 11

1. Find f (x) such that

∫ x

−2f (t) dt = cos(2x) + 1.

2. Give the function f (x) that solves the equation∫ x

2(t + 1)f (t) dt = sin(x).

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Properties

d

dx

(∫ u

af (t) dt

)= f (u)

du

dx

d

dx

(∫ b

vf (t) dt

)= −f (v)

dv

dx

d

dx

(∫ u

vf (t) dt

)= f (u)

du

dx− f (v)

dv

dx

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Problem 12

Find

1.d

dx

(∫ x3

0

dt

1 + t

)

2.d

dx

(∫ x2

−3

(3t − sin(t2)

)dt

)

3.d

dx

(∫ 2x

x

dt

1 + t2

)

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Mean-Value Theorems for Integrals

Let favg denote the average or mean value of f on [a, b]. Then

favg =1

b − a

∫ b

af (x) dx .

The First Mean-Value Theorems for Integrals

If f is continous on [a, b], then there is at least one number c in(a, b) for which

f (c) = favg.

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Problem 13

Give the average value of the function f (x) = sin x on the interval[0, π/2].

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Problem 14

Give the value of c that satisfies the conclusion of the mean valuetheorem for integrals for the function f (x) = x2 − 2x + 3 on theinterval [1, 4].

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Area by Integration with Respect to x : f (x) ≥ g(x)

Rectangle Area[f (x∗i )− g(x∗i )]∆xi

Riemann Sum∑[f (x∗i )− g(x∗i )]∆xi

area(Ω) =

∫ b

a[f (x)− g(x)] dx = lim

‖P‖→0

∑[f (x∗i )− g(x∗i )]∆xi .

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Area by Integration with Respect to y : F (y) ≥ G (y)

Rectangle Area[F (y∗i )− G (y∗i )]∆yi

Riemann Sum∑[F (y∗i )− G (y∗i )]∆yi

area(Ω) =

∫ d

c[F (y)− G (y)] dy = lim

‖P‖→0

∑[F (y∗i )− G (y∗i )]∆yi .

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Problem 15

Give a formula involving integral(s) in x for the region bounded byy = x − 2 and y =

√x .

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Problem 16

Give a formula involving integral(s) in y for the region bounded byy = x − 2 and y =

√x .

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Solid of Revolution About the x-Axis: Washer

Cylinder Volume: π([f (x∗i )]2 − [g(x∗i )]2)∆xi

Riemann Sum:∑

π([f (x∗i )]2 − [g(x∗i )]2)∆xi

V =

∫ b

aπ([f (x)]2−[g(x)]2) dx = lim

‖P‖→0

∑π([f (x∗i )]2−[g(x∗i )]2)∆xi

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Solid of Revolution About the x-Axis: Shell

V =

∫ d

c2π y [F (y)−G (y)] dy = lim

‖P‖→0

∑2π y∗i [F (y∗i )−G (y∗i )]∆yi .

The integrand 2π y [F (y)−G (y)] is the lateral area of the cylinder.

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Problem 17

The region bounded by y = x2 and y = 2x is rotated around thex-axis. Give a formula involving integrals in x for the volume ofthe solid that is generated.

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Problem 18

The region bounded by y = x2 and y = 2x is rotated around thex-axis. Give a formula involving integrals in y for the volume ofthe solid that is generated.

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Solid of Revolution About the y -Axis: Washer

Cylinder Volume: π([F (y∗i )]2 − [G (y∗i )]2)∆yi

Riemann Sum:∑

π([F (y∗i )]2 − [G (y∗i )]2)∆yi

V =

∫ d

cπ([F (y)]2−[G (y)]2) dy = lim

‖P‖→0

∑π([F (y∗i )]2−[G (y∗i )]2)∆yi

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Solid of Revolution About the y -Axis: Shell

V =

∫ b

a2π x [f (x)−g(x)] dx = lim

‖P‖→0

∑2π x∗i [f (x∗i )−g(x∗i )]∆xi .

The integrand 2π x [f (x)− g(x)] is the lateral area of the cylinder.

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Problem 19

The region bounded by y = x2 and y = 2x is rotated around they -axis. Give a formula involving integrals in y for the volume ofthe solid that is generated.

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Problem 20

The region bounded by y = x2 and y = 2x is rotated around they -axis. Give a formula involving integrals in x for the volume ofthe solid that is generated.

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