Sample examinations Calculus II (201-NYB-05) Winter 2009

[Pages:2]Sample examinations

Calculus II (201-NYB-05)

Winter 2009

1. Evaluate

d

n

sin`arccos 1 -

o x2 ? , and simplify

your

answer.

dx

2. Evaluate the following integrals.

a.

Z

2

4 +2 x2 - 1

dx

1

x x2 - 1

Z c. e-2x cos 6x dx

b. Z 5 x + 2 dx 1 2x - 1

Z

d.

t + 1 log t + 1 dt

e.

Z

1 2

sin3

x

cos3

x

dx

Z f.

dx

0

x2 x2 - 36

Z x+4 g. x(x2 + 2) dx

3. Evaluate the following improper integrals.

Z 1 a. 2 1 - x2 dx

Z2 x b. 0 x2 - 4 dx

4. Evaluate the following limits.

a.

lim

x0+

(log x)2 1 + x-1

b. lim (sec x)cot2 x

x0

c.

lim

x

x2 + 2 x-3

-

(x - 2)3 x2 + 1

ff

5. Find the area of the region (in quadrant I) bounded by the graphs of

y

=

2 ,

x

y

=

3x x2 +

2

and

x = 1.

Give the exact answer in simplified form only: no decimals.

6. Let R be the region bounded by the graphs of

y

=

x2 ,

y = x3 - 3x + 3,

x = -2

and

x = 2.

4

a. Set up, but do not evaluate, an integral that represents the volume of the solid

generated by revolving R about the x-axis.

b. Find the volume of the solid generated by revolving R about the line x = 3.

Give the exact answer in simplified form only: no decimals.

7. Find a solution to the differential equation

y

=

p1 - y2 1 + x2 ;

y(1) = 0.

8.

X Let an

n=1

be a series whose nth

partial

sum is given by sn

=

2n + 1 .

n+2

X a. Evaluate an.

b. Find a5.

n=1

9. What can you say about the convergence of each series based only on the limit

of its general term?

X

cos n

a.

n=1 n

b.

1 2

+

1+

1 4

+

1

+

1 8

+1+

1 16

+

1+

?

?

?

10. Determine whether each of the following series converges or diverges; if it

converges, find the sum. Justify your answers.

X

5(-4)n+2

a.

n=1

32n+1

b.

X

log

2n

-

1

n=1 2n + 1

11. Determine whether each of the following series converges or diverges. State

the tests you use and verify that the conditions for using them are satisfied.

X

(n!)2

a.

n=1 (2n)!

X

cos2 k

b.

k=1 k k

X

en

c.

n=1 n

X

,,2?

d. sin

n=2

n

12. Determine whether each of the following series converges absolutely, condi-

tionally or diverges. Justify your answers.

,, X

-n

?3n

a.

n=1 2n + 1

b.

X

(-1)n

log

n

n=2

n

13. Find the radius and interval of convergence for the power series

X

3n (x - 2)n+1.

n=1 2n + 1

14.

Find

the

Taylor

series

of

f (x)

=

cos 2x

centred

at

1 2

.

State

the

first four

non-zero terms and give the formula for the nth term.

Solution outlines

1. sin`arccos 1 - x2 ? = |x|, and hence

d

n

sin`arccos 1

- x2

o ?

=

x

dx

|x|

(1 if x > 0 and -1 if x < 0).

2. a. Separating terms and simplifying reveals two basic integrals:

Z

2

4 +2 x2

-1

dx

=

(4 arcsec x +

2 2 log x)

=

+

log 2.

1

x x2 - 1

1

b. Partial integration gives

Z5

1

x + 2 2x - 1

dx

=

n (x

+

2) 2x

-

1

-

1 3

(2x

-

o5

1)

1

=

1 3

(x

+

7) 2x

-

5 1

1

=

28 3

.

c. By the product rule for differentiation,

d e-2x cos 6x? = -2e-2x cos 6x - 6e-2x sin x, and dx d e-2x sin 6x? = 6e-2x cos 6x - 2e-2x sin x. dx

Subtracting the first equation from three times the second, and then rearranging the corresponding integral equation, one obtains

Z

e-2x

cos 6x dx

=

1 20

e-2x

(3

sin

6x

-

cos 6x)

+

C.

d. Integrate by parts after revising the logarithmic factor, to obtain

Z

1 2

t + 1 log(t + 1) dt =

1 3

(t

+

1)3/2

log(t +

1) -

2 9

(t

+ 1)3/2

+C

=

1 9

`3

log(t

+ 1) - 2?(t + 1)3

+ C.

e. Letting t = sin x gives

f.

Z

1 2

0

Let t =

sin3 x cos3 x dx

x2 - 36/x, or

= t2

Z1 t3(1 - t2) dt =

0

= 1 - 36/x2, so that

1 12

t4

(3

1 36

t

dt

-

=

1 2t2 )

0

dx/x3

=

1 12

.

, and hence

Z

dx

Z =

x2 x2 - 36

x x2 - 36

?

dx x3

=

1 36

Z

dt =

x2 - 36 + C. 36x

g. Resolve the integrand into partial fractions and split the last fraction to integrate.

Z

x+4 x(x2 + 2)

dx

=

Z

2 x

-

2x - 1 ff x2 + 2 dx

=

Z

2 x

-

2x x2 + 2

+

1ff x2 + 2 dx

=

log

x2 x2 +

2

+

1 2

2

arctan

1 2

x2

+

C.

3. a.

Z

2

dx 1 - x2

Zt = lim

t 2

dx 1 - x2

=

lim

t

1 2

log t + 1 t-1

-

1 2

log

3

=

-

1 2

log

3

b. The integral diverges, because

Z2

0

x dx x2 - 4

=

Z lim

2- 0

x dx x2 - 4

=

lim

2-

1 2

log(4 - 2) - log 2

=

-.

4. a. If y = 1/x, then

lim

x0+

(log x)2 1 + x-1

= lim (log y)2 y 1 + y

= lim 2 y y

= 0,

with two applications of l'Ho^pital's rule.

Sample examinations (solution outlines)

Calculus II (201-NYB-05)

Winter 2009

b.

We have

lim (sec x)cot2 x

=

lim

e- cos2 x(log cos x)/ sin2 x

=

e,

since

x0

x0

lim

x0

log cos x sin2 x

=

lim

x0

-1 2 cos2 x

=

-

1 2

,

by l'Ho^pital's rule.

c. Combining terms, and extracting the dominant powers of x from the numerator

and denominator, yields

lim x2 + 2 x x - 3

-

(x - 2)3 x2 + 1

ff

=

lim

x

9 - 17/x + 44/x2 - 24/x3 (1 - 3/x)(1 + 1/x2)

= 9.

5. Since 2/x - 3x(x2 + 2) = (4 - x2)/`x(x2 + 2)? is equal to zero if x = ?2

and positive if 1 x < 2, the area in question is equal to

Z 2 2 1x

-

3x ff x2 + 2 dx

=

1 2

log

x4

2

(x2

+

2)3

1

=

1 2

log 2.

6.

a.

The solid in question can be decomposed into annuli of inner radius

1 4

x2

and

outer radius x3 - 3x + 3, for -2 x 2, so its volume is equal to

2

2

Z

Z

(x3

-

3x

+

3)2

-

(

1 4

x2 )2 ?

dx

=

2x6

-

97 8

x4

+ 18x2

+ 18? dx

-2

0

2

=

2 7

x7

-

97 40

x5

+

6x3

+

18x? 0

=

1504 35

.

b. The solid in question can be decomposed into cylindrical shells of radius 3 - x

and

height

x3

-

1 4

x2

-

3x

+

3,

for

-2

x

2, so its volume is equal to

Z2

Z2

2

(3

-2

-

x)(x3

-

1 4

x2

-

3x

+

3) dx

=

2

0

-2x4 +

9 2

x2

+

18?

dx

2

=

2

2 5

x5

+

3 2

x3

+ 18x? 0

=

352 5

.

7. Separating variables gives

Z

dy

Z =

p1 - y2

dx 1 + x2 ,

or

arcsin y = arctan x + C.

Since

y(1)

=

0

implies

that

C

=

1 4

,

we

must

have

x

-1, in which case

y

=

sin(arctan x -

1 4

)

=

x-1 p2(x2 +

. 1)

8.

a.

X The sum of the series is an

n=1

=

lim

n

sn

=

lim

n

2n + 1 n+2

=

2.

b. The series begins with n = 1, so sn = a1 + ? ? ? + an (n terms), and therefore

a5

=

s5

- s4

=

11 7

-

3 2

=

1 14

.

9. a. Since -1/n < (cos n)/n < 1/n for n 1, and lim (?1/n) = 0,

n

lim (cos n)/n = 0 by the Squeeze Theorem. In this case, no conclusion can be

n

drawn about the series only from the limit of its general term.

b. Among infinitely many possibilities:

?

if an

=

2-

1 2

(n+1)

sin2

1 2

n

+

cos2

1 2

n

for

n

1, then a2n = 1 for

?

n if

an

1=an(d|nelim-na|n+=e0,-son|P)e(-a2nnnd|+iv1e2crngo+ess23b12ythne+vasniins2hin21gcnr)itefroironn;

1,

then

lim

n

an

=

0 (since

an

=

0 for n

9) and no conclusion can be

drawn about the series only from the limit of its general term;

?

if an

=

4 XY

i=1 1 j 8

2-i(n - j) 2i - j - 1

+

Y

n-j ff for n

1 j 8 2i - j

j=2i-1

j=2i

lim

n

an

=

and

P

an

diverges

by

the

vanishing

criterion.

No content in sight--just a matter of "guess what teacher wants to hear."

1, then

10.

a.

This

is

a

geometric

series

with

first

term

-

320 27

and

common

ratio

-

4 9

,

so

X

5(-4)n+2

n=1 32n+1

=

-320/27 1 - (-4/9)

=

-

320 39

.

b. This is a divergent telescoping series, since

X

log

2n

-1

=

lim

n

X log

2k

-1

=

-

lim

log(2n + 1) = -.

n=1

2n + 1

n k=1

2k + 1

n

11. a. Let n = (n!)2/(2n!); then

lim

n+1

n n

=

lim

n

n+1 2(2n + 1)

=

1 4

,

which is smaller than one. Therefore, P n converges by the ratio test.

b. Since

0

<

cos2 k

<

1

kk kk

for k 1,

X

cos2 k

k=1 k k

converges with the p-series P k-3/2 by the comparison test.

c. Since ex > x for all real numbers x, the series

X

en

n=1 n

diverges by the vanishing criterion.

d.

Since

1 2

<

cos

<

sin if 0

<

<

1 3

,

1/n

<

sin(2/n) if n

so

X

,, sin

2

?

n=2

n

diverges with the harmonic series by the comparison test.

2, and

12. a. Let n = `-n/(2n + 1)?3n; then

lim

n

n |n|

=

,, 1 ?3 lim n 2 + 1/n

=

1 8

,

which b. Let

is f

less than one. Therefore, (x) = (log x)/x; then

P n is absolutely f is decreasing on (

convergent by e2, ), since

the

root

test.

f (x) = (2 - log x)/(2xx) < 0

if x > e2, f (x) > 0 if x > 1, and f (x) 0 as x by (one application of)

l'Ho^pital's rule. Therefore, On the other hand, f (x) >

P1/(-x1)ifnxf

(n) converges by > e, so P f (n)

the alternating series test. diverges with the p-series

P n-1/2 by the comparison test. Hence, the series

X

(-1)n

log

n

n=2

n

is conditionally convergent.

13. Let n = 3n(x - 2)n+1/(2n + 1); then

lim

n+1

=

lim

3(2n

+

1) |x

-

2|

=

3|x

-

2|,

n n n 2n + 3

so by the ratio test P n

converges absolutely if

5 3

<

x

<

7 3

and diverges if

x

<

5 3

or x

>

7 3

.

If

x

=

5 3

then

n

=

(-1)n+1 /(2n

+ 1),

so

P n

converges

by the alternating series test (since 1/(2n + 1)? is positive and decreasing, and

its

limit

is

zero).

If

x

=

7 3

then

n

=

1/(2n

+

1)

1/(3n) if n

1, so P n

diverges with the harmonic series by the comparison test. Therefore, the interval

of

convergence

of

P n

is

^

5 3

,

7 3

);

its

radius

of

convergence

is

1 3

.

14. Using the Maclaurin series of cos t, where t = 2x - , gives

cos 2x

=

-

cos(2x

-

)

=

-

cos

2`x

-

1 2

?

=

X

k=0

(-1)k+1 22k (2k)!

`x

-

1 2

?2k

=

-1 + 2`x

-

1 2

?2

-

2 3

`x

-

1 2

?4

+

4 45

`x

-

1 2

?6

-

...

2

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