Exam 1 Study Guide. Differentiation and Anti-differentiation Rules from Calculus I

Exam 1 Study Guide

Math 2020 - Calculus II, Winter 2014

The following is a list of important concepts from each section that will be tested on exam 1. This is not a complete list of the material that you should know for the course, but it is a good indication of what will be emphasized on exam. A thorough understanding of all of the following concepts will help you perform well on exam 1. Some places to find problems on these topics are the following: in the book, in the homework, on quizzes, and online (for example the COW webpage).

Differentiation and Anti-differentiation Rules from Calculus I

d dx

(c) = 0

d dx

(e

x

_{) = e}

x d dx

ln(x) =

1 x d dx

(x

n

_{) = n x}

n−1 d dx

(a

x

_{) = a}

x

_ln(a)

d dx

log

a

(x) =

_xln(a)1 d dx

sin(x) = cos(x)

d dx

cos(x) = − sin(x)

d dx

tan(x) = sec

2

(x)

d dx

csc(x) = − csc(x) cot(x)

d

dx

sec(x) = sec(x) tan(x)

d dx

cot(x) = − csc

2

(x)

d dx

(sin

−1

_{(x)) =}

_√1 1−x2 d dx

(cos

−1

_{(x)) = −}

_√1 1−x2 d dx

(tan

−1

_{(x)) =}

1 1+x2 d dx

(csc

−1

_{(x)) = −}

1 x√x2₋₁ d dx

(sec

−1

_{(x)) =}

1 x√x2₋₁ d dx

(cot

−1

_{(x)) = −}

1 1+x2. Z

k dx

= k x +C

Z

e

x

dx

= e

x

+C

Z

₁

x

dx

= ln(|x|) +C

Z

x

n

dx

=

1

n

+ 1

n x

n+1

_+C

;

n

6= −1

Z

a

x

dx

=

a

x

ln(a)

+C

Z

cos(x)dx = sin(x) +C

Z

sin(x)dx = − cos(x) +C

Z

sec

2

(x)dx = tan(x) +C

Z

csc(x) cot(x)dx = − csc(x) +C

Z

sec(x) tan(x)dx = sec(x) +C

Z

csc

2

(x) = − cot(x) +C

Z

₁

√

1 − x

2

dx

= sin

−1

_{(x) +C}

Z

_√

1

1 − x

2

dx

= − cos

−1

_{(x) +C}

Z

1

1 + x

2

= tan

−1

_{(x) +C}

Z

₁

x

√

x

2

_{− 1}

dx

= − csc

−1

_{(x) +C}

Z

1

x

√

x

2

_{− 1}

dx

= sec

−1

_{(x) +C}

Z

1

1 + x

2

dx

= − cot

−1

_{(x) +C}

_.

Techniques of Integration

Section 5.2 This section is on computing definite integrals as limits of Riemann sums and in terms of signed

area.

•

Interpret a definite integral as signed area.

•

Compute integrals using area formulas of of geometric figures.

Section 5.3 This section is on the fundamental theorem of calculus.

•

Use the fundamental theorem of calculus to compute definite integrals, i.e. Z b

a

f

(x)dx = F(b) − F(a),

where

F(x)

is an anti-derivative of

f

(x)

on

[a, b]

.

•

Compute the indefinite integrals from the table above Z

f

(x)dx = F(x) +C,

where

F(x)

is an anti-derivative of

f

(x)

.

Section 5.5 This section is on

u

-substitution.

• u

-substitution is the anti-differentiation rule formulated from “undoing” the chain rule:

Chain Rule

u

-substitution

d

dx

f(g(x)) = f

0

(g(x))g

0

(x)

←→

Z

f

0

(g(x))g

0

(x)dx = f (g(x)) +C

•

Use

u

-substitution to evaluate indefinite integrals

Z

f

0

(g(x))g

0

(x)dx =

Z

f

0

(u)du = f (u) +C = f (g(x)) +C

u

-substitution



u

= g(x)

du

= g

0

(x)dx



•

Use

u

-substitution to evaluate definite integrals Z b a

f

0

(g(x))g

0

(x)dx =

Z g(b) g(a)

f

0

(u)du = f (g(a)) − f (g(b))

u

-substitution



u

= g(x)

u(a) = g(a)

du

= g

0

(x)dx

u(b) = g(b)



•

Integration by parts is the anti-differentiation rule formulated from “undoing” the product rule:

Product Rule Integration by parts

d

dx

( f (x) · g(x)) = g(x) f

0

(x) + f (x)g

0

(x)

←→

Z

f

(x)g

0

(x)dx = f (x)g(x) −

Z

g(x) f

0

(x)dx

•

Use integration by parts to evaluate indefinite integrals

Z

f

(x)g

0

(x)dx = f (x)g(x) −

Z

g(x) f

0

(x)dx

Integration by parts



u

= f (x)

v

= g(x)

du

= f

0

(x)dx

dv

= g

0

(x)dx



This formula says thatR

u dv

= u · v −

R

v du

. When integrating by parts, we choose

u

and

dv

must then be the remainder of the integrand. Then

du

is the differential of

u

and

v

is chosen by anti-differentiating

g

0

(x)

so the

dv

is the differential of

v

.

•

Use integration by parts to evaluate definite integrals Z b a

f

(x)g

0

(x)dx = f (x)g(x)









b a

−

Z b a

g(x) f

0

(x)dx

Integration by parts



u

= f (x)

v

= g(x)

du

= f

0

(x)dx

dv

= g

0

(x)dx



The techniques here are the same as integrating by parts to evaluate indefinite integrals. The difference is that we evaluate the terms at the endpoints

a

and

b

.

Section 7.2 This section is on techniques for integrating trigonometric functions.

•

Compute indefinite integrals of the form Z

sin

m

(x) cos

n

(x)dx

when either

m

or

n

is odd by “splitting off a derivative term,” using the Pythagorean theorem, and integrate using

u

-substitution. You should be able to compute definite integrals of this form as well. You are expected to know the Pythagorean theorem,

sin

2

(x) + cos

2

(x) = 1.

•

Compute indefinite integrals of the form

Z

sin

m

(x) cos

n

(x)dx

when both

m

or

n

are even by using double angle formulas. The double angle formulas,

cos

2

(x) =

1

2

+

1

2

cos(2x)

and

sin

2

_{(x) =}

1

2

−

1

2

cos(2x)

will be provided for you; you are not expected to memorize them. You should be able to compute definite integrals of this form as well.

•

Compute indefinite integrals of the form Z

tan

m

(x) sec

2k

(x)dx

and

Z

tan

2k+1

(x) sec

n

(x)dx

by “splitting off a derivative term,” using the Pythagorean theorem, and integrate using

u

-substitution. You should be able to compute definite integrals of this form as well. You are expected to know another version of the Pythagorean theorem, obtained in the following way: start with the Pythagorean theorem,

sin

2

(x) + cos

2

(x) = 1

, and divide both side by

cos

2

(x)

to obtain

sin

2

(x)

cos

2

_(x)

+

cos

2

(x)

cos

2

_(x)

=

1

cos

2

_(x)

tan

2

(x) + 1 = sec

2

(x).

Section 7.4 This section is on integrating rational functions using partial fraction decomposition.

•

Use partial fraction decomposition to rewrite rational functions into a sum of functions that you can integrate directly. This may involve polynomial long division (or synthetic division if you prefer).

•

You will be expected to be able to decompose rational functions whose denominators have linear terms, repeated linear terms, irreducible quadratic terms, and combinations thereof. You will not be expected to handle repeated irreducible quadratic terms. In particular, the denominators of the functions you will be expected to integrate will contain terms of the following forms:

Linear term Repeated linear term Irreducible quadratic term

(x − a)

(x − a)

k

(x

2

+ a

2

)

For example, you are expected to be able to integrate

x

4

(x − 1)(x − 3)

3

_(x

2

_{+ 1)}

.

This particular problem would be too long for an exam question, but you should know all of the tech-niques involved. You should be able to compute both definite and indefinite integrals of this form.

Applications of Integration

Section 6.1 This section is on computing area between curves.

•

Given two functions

f

(x)

and

g(x)

on an interval

[a, b]

, you should be able to compute the area between the two curves on

[a, b]

A

=

Z b a

| f (x) − g(x)|dx.

In practice, we break up the interval

[a, b]

in to sections where

f

(x)

is above

g(x)

and vice versa, then integrate the top function minus the bottom function on each interval.

•

Given two functions

f

(y)

and

g(y)

on an interval

[c, d]

, you should be able to compute the area between the two curves on

[c, d]

A

=

Z d c

| f (y) − g(y)|dy.

This is essentially the same problem as the last bullet item turned on its side. We do everything in the

y

direction instead of the

x

direction.

•

You should be able to identify regions defined by their bounding curves. That is, you should be able to identify a region given the curve that form its boundary. For example, you should be able to identify the region bounded by

y

= sin(x)

,

y

= −1

,

x

= −π/2

, and

x

= 3π/2

.

Section 6.2 This section is on computing volumes of solids.

•

You should be able to use the formula

V

=

Z b a

A(x)dx

where

V

is the volume of the solid with cross-sectional area

A(x)

for

x

in

[a, b]

. You should be able to use this formula when a solid is described by its base and cross-sectional region as well as using revolutions of solids.

•

Use the above formula (using the disk or washer method) to compute the volume of solids of revolution obtained by revolving a given region about a horizontal or vertical axis. There are two formulations of this formula to compute volumes.

Revolving about a horizontal axis Revolving about a vertical axis -Axis of revolution is of the form

y

= a

-Axis of revolution is of the form

x

= a

(including the

x

-axis, which is

y

= 0

) (including the

y

-axis, which is

x

= 0

) -Vertical cross-sectional cuts -Horizontal cross-sectional cuts

(parallel to the

y

-axis) (parallel to the

x

-axis)

-Fix

x

and compute cross-sectional area

A(x)

-Fix

y

and compute cross-sectional area

A(y)

-Integrate in

x

-Integrate in

y

-Integrating with disks: -Integrating with disks:

V

=

Z b a

πR(x)

2

dx

V

=

Z d c

πR(y)

2

dy

-Integrating with washers: -Integrating with washers:

V

=

Z b a

π(R(x)

2

− r(x)

2

)dx

V

=

Z d c

π(R(y)

2

− r(y)

2

)dy

Section 6.4 This section is on computing work.

•

You should be familiar with the formulas

W

= F · d

(work = force

×

distance) and

F

= m · a

(force = mass

×

a)

•

You should be able to set up and compute integrals to find the amount of work done in various appli-cations problems like the ones in section 6.4 in the text. This includes using Hooke’s law to compute the force function for a spring, computing work given a force function, work done to pump water out of a tank, work done to lift a chain/rope, etc.

Section 6.5 This section is on computing average values of functions.

•

Given a function

f(x)

on

[a, b]

, the average value of

f

(x)

on

[a, b]

is

f

_[a,b]

=

1

b

− a

Z b a

f

(x)dx.

You should know this equation and be able to compute integral averages using the integration tech-niques mentioned above.

•

Know the interpretation of the average value of a function in terms of the area under a curve. In particular,

f

_[a,b] is the height of the rectangle of width

b

− a

necessary to enclose the same area underneath

f

(x)

on

[a, b]

. That is,

f

_[a,b]

· (b − a) =

Z b a

f

(x)dx



Area of a rectangle with height

f

_[a,b]and width

b

− a

.



=



Area under

f

(x)

on

[a, b]

.