M53 E4 Reviewer.pdf

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(1)

Unit IV Exam

Reviewer

(2)

Evaluate

Z

cot

x

(

cot

x

tan

x

)

dx

Z

cot

x

(

cot

x

tan

x

)

dx

=

Z

(

cot

2

x

1

)

dx

=

Z

(

csc

2

x

1

1

)

dx

=

Z

(

csc

2

x

2

)

dx

(3)

Evaluate

Z

x

5

(

x

3

1

)

8

dx

Let

u

=

x

3

1

. This implies that

(1)

du

=

3

x

2

dx

or

1

3

du

=

x

2

dx

and

(2)

x

3

=

u

+

1

.

Z

x

5

(

x

3

1

)

8

dx

=

Z

x

3

(

x

3

1

)

8

x

2

dx

=

Z

(

u

+

1

)

u

8

1

3

du

=

1

3

Z

(

u

9

+

u

8

)

du

=

1

3

u

10

10

+

u

9

9

+

C

=

1

3

(

x

3

1

)

10

10

+

(

x

3

1

)

9

9

(4)

Evaluate

Z

1

0

(

|

2

3

x

|

+

1

)

dx

We first take note that

|2

3

x

|

=

2

3

x x

2

3

−2

+

3

x x

>

2

3

Z

1

0

(

|2

3

x

|

+

1

)

dx

=

Z

2

3

0

(

2

3

x

+

1

)

dx

+

Z

1

2 3

(

−2

+

3

x

+

1

)

dx

=

Z

2

3

0

(

3

3

x

)

dx

+

Z

1

2 3

(

−1

+

3

x

)

dx

=

3

x

3

x

2

2

2 3

0

+

x

+

3

x

2

2

1

2 3

=

2

2

3

0

+

1

2

0

=

11

(5)

Evaluate

Z

π2

/

16

π2

/

36

sec

2

x

x

dx

Let

u

=

x. This implies

du

=

1

2

x

dx. Also,

x

=

π

2

36

−→

u

=

π

6

and

x

=

π

2

16

−→

u

=

π

4

Z

π2

/

16

π2

/

36

sec

2

x

x

dx

=

Z

π

/

4

π

/

6

2 sec

2

u du

=

2 tan

u

π

/

4

π

/

6

=

2 tan

π

4

2 tan

π

6

=

2

2

(6)

Find

b

such that the average value of

h

(

x

) =

3

x

2

on

[

0,

b

]

is equal to 1.

Since

h

(

x

) =

3

x

2

is continuous in

R

, then

h

is also continuous on

[

0,

b

]

.

By the MVT for integrals,

have

=

Z

b

0

3

x

2

dx

b

0

=

x

3

b

0

b

=

b

3

b

=

b

2

(7)

Given

G

(

x

) =

Z

tan

x

4 π

x

cos

(t

1

)

t

dt

. Find:

(a)

G(

π

/4

)

(b)

G

0

(

π

/4

)

(a)

G

(

π

/4

) =

Z

1

1

cos

(

t

1

)

t

dt

=

0

(b)

G

(

x

) =

Z

1

4 π

x

cos

(

t

1

)

t

dt

+

Z

tan

x

1

cos

(

t

1

)

t

dt

G

(

x

) =

Z

4

π

x

1

cos

(

t

1

)

t

dt

+

Z

tan

x

1

cos

(

t

1

)

t

dt

G

0

(

x

) =

cos

(

4x

π

1

)

4x

π

4

π

+

cos

(

tan

x

1

)

tan

x

sec

2

x

(8)

Elsa threw her crown upwards with an initial velocity of 24

ft

/

s

, from a

height of 40

ft

. From the same height at the same time, she released her

cape and just let it go. Assume that no other forces except acceleration due

gravity of -32

ft

/

s

2

affects the crown and cape.

1

What is the maximum height of the crown?

2

Which item is moving faster at time

t

=

1

. Explain your answer.

acrown

(

t

)

=

−32

vcrown

(

t

)

=

−32

t

+

C

1

vcrown

(

0

)

=

C

1

=

24

vcrown

(

t

)

=

−32

t

+

24

scrown

(

t

)

=

−16

t

2

+

24

t

+

C

2

scrown

(

0

)

=

C

2

=

40

scrown

(

t

)

=

−16

t

2

+

24

t

+

40

(9)

Elsa threw her crown upwards with an initial velocity of 24

ft

/

s

, from a

height of 40

ft

. From the same height at the same time, she released her

cape and just let it go. Assume that no other forces except acceleration due

gravity of -32

ft

/

s

2

affects the crown and cape.

1

What is the maximum height of the crown?

2

Which item is moving faster at time

t

=

1

. Explain your answer.

acape

(

t

)

=

−32

vcape

(

t

)

=

−32

t

+

C

1

vcape

(

0

)

=

C

1

=

0

vcape

(

t

)

=

−32

t

scape

(

t

)

=

−16

t

2

+

C

2

scape

(

0

)

=

C

2

=

40

scape

(

t

)

=

16

t

2

+

40

(10)

x

y

R

(

−3, 0

)

(

−1, 0

)

(

1, 4

)

y

=

2

x

3

+

2

y

=

x

+

3

Let

R

be the region bounded by the

graphs

y

=

2

x

3

+

2

,

y

=

x

+

3

and

the

x-axis. Set-up the integral

1.

the definite integral for the area

of

R

2.

the definite integral for the

perimeter of

R

3.

the definite integral for the

volume of the solid generated

when

R

is revolved about

(11)

x

y

R

(

−3, 0

)

(

−1, 0

)

(

1, 4

)

y

=

2

x

3

+

2

y

=

x

+

3

Area of

R

: The best method uses

horizontal strips. So our integral is

in terms of

y.

A

R

=

Z

4

0

3

r

y

2

2

(

y

3

)

!

dy

If we use vertical strips,

A

R

=

Z

1

3

((

x

+

3

)

0

)

dx

+

Z

1

1

(12)

x

y

R

(

−3, 0

)

(

−1, 0

)

(

1, 4

)

y

=

2

x

3

+

2

y

=

x

+

3

Perimeter of

R

: We set-up with

respect to

x.

L

R

= (

−1

(

−3

))

+

Z

1

3

q

1

+ (

1

)

2

dx

+

Z

1

1

q

(13)

x

y

y

=

0

R

(

−3, 0

)

(

−1, 0

)

(

1, 4

)

y

=

2

x

3

+

2

y

=

x

+

3

Volume of the SOR when

R

is

revolved about

y

=

0

using

cylindrical shells

V

=

2

π

r h w

V

R

=

Z

4

0

y

3

r

y

2

2

(

y

3

)

!

(14)

x

y

x

=

1

R

(

−3, 0

)

(

−1, 0

)

(

1, 4

)

y

=

2

x

3

+

2

y

=

x

+

3

Volume of the SOR when

R

is

revolved about

x

=

1

using washers

V

=

π

(

r

2

2

r

2

1

)

h

V

R

=

Z

4

0

π

1

(

y

3

)

2

1

q

3

y

2

2

2

#

(15)

More items for review

I.

Evaluate the following integrals.

1.

Z

x

5

(

x

3

+

1

)

4

dx

2.

Z

sin

x

(

1

+

sin

x

cos

x

)

dx

3.

Z

x

csc

2

(

tan

x

2

)

cos

2

(

x

2

)

dx

4.

Z

0

π

/

3

sec

x

tan

x

3

2

sec

x dx

II.

Let

R

be the region bounded by the graphs of

x

=

1

y

2

,

x

+

y

+

1

=

0

, and the

x-axis. Set-up the definite integral(s) for

1

the area of

R

.

2

the perimeter of

R

.

(16)

Figure

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References

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Related subjects : ft/s