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Answers to Exam Problems: Integration and Series, Exams of Calculus

Solutions to practice problems on integration and series, including using techniques such as partial fractions, reduction formulas, and the comparison test. Topics covered include improper integrals, definite integrals, and infinite series.

Typology: Exams

Pre 2010

Uploaded on 08/05/2009

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Answers to Practice Problems Exam 3
1. a)
Let x=1sinθ, so dx = cos(θ). Then (1 x
2
)
3
2
dx =cos(θ)
(1 sin
2
(θ))
3
2
=cos(θ)
(cos
2
(θ))
3
2
=cos(θ)
cos
3
(θ)
=1
cos
2
(θ)
= sec
2
(θ)
= tan(θ) + C.
Since x=1sinθ, then x
1= sinθ = opposite
hypot. . Label triangle with x= opposite
and 1= hypot. The missing side (adjacent) will have length 1 x2.
Thus, tanθ=
opposite
adjacent
=
x
1−x
2
, so (1 x
2
)
3
2
dx =
x
1−x
2
+C.
b)
Let x= 4 tanθ, so dx = 4 sec
2
(θ). Then dx
x
2
+16
=4 sec
2
(θ)
(16tan
2
(θ) +16)
1
2
=4 sec
2
(θ)
(16(tan
2
(θ) +1))
1
2
=4 sec
2
(θ)
(16sec
2
(θ))
1
2
=4 sec
2
(θ)
4 sec(θ)
= sec(θ)
=lnsec(θ) + tan(θ) + C. Since x= 4 tanθ, then
x
4
= tanθ =
opposite
adjacent
.
Label triangle with x= opposite and 4 = adjacent. The missing side
(hypot.) will have length x
2
+16. Then sec(θ) =
hypot.
adjacent
=
x
2
+16
4
and tan(θ) =
x
4
,
so dx
x
2
+16
= ln
x
2
+16
4
+
x
4
+C.
c)
sin
3
xcos
3
xdx = (sin
3
x)(cos
2
x)(cos x)dx = (sin
3
x)(1 sin
2
x)(cos x)dx
d)
tan xcsc xdx
=sin(x)
cos(x)
1
sin(x)dx =1
cos(x)
dx = sec(x)
dx = ln sec(x) + tan( x) + C.
e)
(cos x+ sin x)
2
dx = cos
2
(x) + 2(cos x)(sin x) + sin
2
(x)
dx
=(sin
2
(x)
+ cos
2
(x)) + 2(cos x)(sin x)dx = 1+ 2(cos x)(sin x)
dx
=x+ 2 (sin x)(cos x)
dx ( Now let u= sin x, so du = cos xdx. )
=x+ 2 udu
=x+ 2
1
2
( )u
2
+C=x+ sin
2
(x) + C.
f)
Let u= 3x, so
1
3
( )du =dx. Then cos
3
(3x
)dx =
1
3
cos
3
(u
)du
1
pf3
pf4
pf5

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Answers to Practice Problems Exam 3

  1. a)

Let x = 1 sin θ , so dx = cos( θ ) . Then ( 1 − x

2

)

3

2

dx =

cos( θ )

( 1 − sin

2

( θ ))

3

2

=

cos( θ )

(cos

2

( θ ))

3

2

=

cos( θ )

cos

3

( θ )

=

1

cos

2

( θ )

= sec

2

( θ )

= tan( θ ) + C.

Since x = 1 sin θ , then

x

1

= sin θ =

opposite

hypot.

. Label triangle with x = opposite

and 1 = hypot. The missing side (adjacent) will have length 1 − x

2

Thus, tan θ =

opposite

adjacent

=

x

1 − x

2

, so ( 1 − x

2

)

3

2

dx =

x

1 − x

2

  • C.

b)

Let x = 4 tan θ , so dx = 4 sec

2

( θ ) . Then

dx

x

2

  • 16

=

4 sec

2

( θ )

( 16 tan

2

( θ ) + 16 )

1

2

=

4 sec

2

( θ )

( 16 (tan

2

( θ ) + 1 ))

1

2

=

4 sec

2

( θ )

( 16 sec

2

( θ ))

1

2

=

4 sec

2

( θ )

4 sec( θ )

= sec( θ )

=lnsec( θ ) + tan( θ ) + C. Since x = 4 tan θ , then

x

4

= tan θ =

opposite

adjacent

.

Label triangle with x = opposite and 4 = adjacent. The missing side

(hypot.) will have length x

2

    1. Then sec( θ ) =

hypot.

adjacent

=

x

2

  • 16

4

and tan( θ ) =

x

4

, so

dx

x

2

  • 16

= ln

x

2

  • 16

4

x

4

  • C.

c)

sin

3

x cos

3

xdx = (sin

3

x )(cos

2

x )(cos x ) dx = (sin

3

x )( 1 − sin

2

x )(cos x ) dx

= ( u

3

)( 1 − u

2

) du ( where u = sin( x ) and du = (cos x ) dx )

= u

3

u

5

du =

1

4

u

4

1

6

u

6

+ C =

1

4

sin

4

( x ) −

1

6

sin

6

( x ) + C.

d)

tan x csc xdx

sin( x )

cos( x )

sin( x )

dx =

cos( x )

dx = sec( x )

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

e)

(cos x + sin x )

2

dx = cos

2

( x ) + 2 (cos x )(sin x ) + sin

2

( x )

dx

= (sin

2

( x )

  • cos

2

( x )) + 2 (cos x )(sin x ) dx = 1 + 2 (cos x )(sin x )

dx

= x + 2 (sin x )(cos x )

dx ( Now let u = sin x , so du = cos xdx. )

= x + 2 udu

= x + 2

1

2

u

2

+ C = x + sin

2

( x ) + C.

f)

Let u = 3 x , so

1

3

( )

du = dx. Then cos

3

( 3 x

) dx =

1

3

cos

3

( u

) du

1

3

cos

2

( u )(sin u )

3

2

3

cos( u

) du

[ ]

(using a reduction formula)

1

3

cos

2

( u )(sin u )

3

2

3

sin( u )

[ ]

+ C =

1

3

cos

2

( 3 x )(sin( 3 x ))

3

2

3

sin( 3 x )

[ ]

+ C.

g)

sin x sin( 2 x ) dx

= sin x ( 2 (sin x )(cos x )) dx = 2 sin

2

( x )(cos x ) dx

= 2 u

2

du =

2

3

u

3

+ C =

2

3

sin

3

( x ) + C.

1. h)

cos

4

xdx

cos

3

( x )(sin x )

4

3

4

cos

2

( x

) dx

[ ]

(using a reduction formula)

cos

3

( x )(sin x )

4

3

4

1

2

( 1 + cos( 2 x )

) dx + C =

cos

3

( x )(sin x )

4

3

8

( x +

1

2

sin( 2 x )) + C.

(or :

cos

3

( x )(sin x )

4

3

8

x +

3

8

1

2

sin( 2 x )) + C =

cos

3

( x )(sin x )

4

3

8

x +

3

8

1

2

( 2 )(sin x )(cos x )) + C

cos

3

( x )(sin x )

4

3

8

x +

3

8

(sin x )(cos x ) + C .)

i)

For

2 x − 3

x ( x − 3 )( x

2

dx , use partial fractions :

2 x − 3

x ( x − 3 )( x

2

A

x

B

x − 3

Cx + D

x

2

Then 2 x − 3 = A ( x − 3 )( x

2

  • 1 ) + Bx ( x

2

  • 1 ) + ( Cx + D ) x ( x − 3 ).

For x = 3 : 3 = A ( 0 ) + B ( 3 )( 10 ) + ( Cx + D )( 0 ), so B =

1

10

For x = 1 : − 1 = A (− 2 )( 2 ) + B ( 1 )( 2 ) + ( C + D )( 1 )(− 2 ),

so − 1 = ( 1 )(− 4 ) + (

1

10

)( 2 ) + ( C + D )(− 2 ) so C +D = −

7

5

For x = 2 : 1 = A (− 1 )( 5 ) + B ( 2 )( 5 ) + ( 2 C + D )( 2 )(− 1 ),

so 1 = ( 1 )(− 5 ) + (

1

10

)( 10 ) + ( 2 C + D )(− 2 ) so 2 C +D = −

5

2

Since C + D = −

7

5

, then - C - D =

7

5

. Using this equation

together with 2 C +D = −

5

2

(add), we find that C = −

11

10

and since C +D = −

7

5

, then D = −

3

10

Thus

2 x − 3

x ( x − 3 )( x

2

  • 1 )

dx =

1

x

1

10

x − 3

11

10

x

3

10

x

2

  • 1

dx

= ln x +

1

10

ln x − 3 +

11

10

x

x

2

  • 1

dx

3

10

x

2

  • 1

dx

= ln x +

1

10

ln x − 3 −

11

10

1

2

( )

1

u

du

3

10

arctan( x ) + C

= ln x +

1

10

ln x − 3 −

11

20

ln x

2

  • 1 −

3

10

arctan( x ) + C.

j)

For

x

4

− x

2

dx , use partial fractions :

x

4

− x

2

x

2

( x

2

x

2

( x + 1 )( x − 1 )

A

x

B

x

2

C

x + 1

D

x − 1

Thus,

dx

x − 2

3

1

29

= −

3

2

27

2

=

24

2

= 12.

dx

x − 2

3

1

29

converges to 12.

c)

x

3

−∞

− 2

dx = lim

t →−∞

2 x

− 1 / 3

x = t

− 2

dx = lim

t →−∞

x = t

x

2 / 3

[ ]

= lim

t →−∞

2 / 3

t

2 / 3

=

3

2

(− 2 )

2 / 3

− ∞ = −∞, so

2

x

3

−∞

− 2

dx diverges.

d)

dx

1 − x

2

0

1

= lim

t → 1

dx

1 − x

2

x = 0

t

= lim

t → 1

x = 0

t

arcsin( x ) [ ]

=lim

t → 1

(arcsin( t ) − arcsin( 0 )) = lim

t → 1

(arcsin( t ) − 0 ) =

π

.

3. e)

dx

x + e

x

0

(Since we can't find the antiderivative of f(x) =

1

x + e

x

,

we will use the Comparison Test for Nonnegative Integrals.)

Thus, x + e

x

≥ 0 + e

x

, so x + e

x

e

x

Hence,

x + e

x

e

x

so

dx

x + e

x

0

e

x

dx =

0

e

x

dx

0

e

x

dx

0

= lim

t →∞

e

x

dx

x = 0

t

= lim

t →∞

x = 0

t

e

x

[ ]

= lim

t →∞

x = 0

t

e

x [ ]

= lim

t →∞

e

t

e

0

= lim

t →∞

e

t

Since

dx

x + e

x

0

≤ e

x

dx

0

and e

x

dx

0

converges to 1,

then by the Comparison Test for Nonnegative Integrals,

dx

x + e

x

0

also converges.

f)

xdx

x

2

  • 2

6

= lim

t →∞

xdx

x

2

  • 2

x = 6

t

= lim

t →∞

1

2

1

u

x = 6

t

du

= lim

t →∞

x = 6

t

1

2

ln u

[ ]

=lim

t →∞

x = 6

t

1

2

ln x

2

  • 2

[ ]

= lim

t →∞

1

2

ln t

2

  • 2 −

1

2

ln( 38 )

( )

= ∞ −

1

2

ln( 38 ) = ∞,

so the integral

xdx

x

2

  • 2

6

diverges.

g)

dx

x

4

  • x

2

(Since we can't find the antiderivative of f(x) =

1

x

4

  • x

,

we will use the Comparison Test for Nonnegative Integrals.)

Since we're integrating from 2 to ∞, x ≥ 2. Thus, x ≥ 0 , so x

4

  • xx

4

Therefore,

x

4

  • x

x

4

, so

dx

x

4

  • x

2

x

4

dx

2

Note that, since

x

4

dx

2

has the form

x

p

dx

a

, where p > 1,

then we know that

x

4

dx

2

converges.

Or, we can prove this using limits:

x

4

dx

2

= lim

t →∞

x

− 4

dx

x = 2

t

= lim

t →∞ x = 2

t

1

3

x

− 3

[ ]

= lim

t →∞

x = 2

t

3 x

3 [ ]

= lim

t →∞

3 t

3