Math 180 Exam III Past Exam Solutions
1) a)

x
dx
( x  3) 6
: let
u  x  3.
Then
du  dx .
Solving for x, we have
x  u  3.
Thus

x
(u  3)
1
3
u 4 3u 5
1
1
dx

du

(

)
du


c 

c
6
6
5
6
4


( x  3)
u
u
u
4 5
4( x  3)
5( x  3)5
b)

1
2
x  1dx :
Break up the integral into two pieces: the left portion of V and the right portion of V.
Thus you need to separate the integral at x = -1.
c) Use change of basis formula:
u  ln x  du 
1
dx .
x
We have
log x
 x2 dx  
1
ln x
dx 

ln 2
x
1

1
1
2
1
x  2 dx    ( x  1)dx   ( x  1)dx 
2
3
2
ln x
ln x
ln 2 dx  1
dx . Next let

x
ln 2
x
1
1 u2
(ln x) 2
udx


c

c
ln 2 
ln 2 2
2 ln 2
d) Use u-sub and an integral formula for the exponential function: Let
1
u  x  1  du  2 xdx  du  xdx .
2
2
e) Use the plus/minus trick:
1
2u
2 x 1
dx   2u du 
c 
c
2
2 ln 2
2 ln 2
2

x2
x 2 1
2x
2x  1  1 2x  1
1
1



 1
.
2x  1
2x  1
2x  1 2x  1
2x  1
Thus
2x
1
dx   (1 
)dx . The second integral can be compute using
2x  1
2x  1
1
1
1 du 1
1
u  2 x  du  2dx  du  dx . And 
dx  
 ln u  c  ln 2 x  1 .
2
2x  1
2 u 2
2
2x
1
1
 2 x  1 dx   (1  2 x  1)dx  x  2 ln 2 x  1  c

f)
 sin


u3 
cos3 x
sin 3 x dx   sin x(1  cos 2 x)dx   (1  u 2 )du   u    c  (cos x 
)c
3
3

3
x dx   sin x(1  cos 2 x)dx .
Therefore,
Let
u  cos x  du   sin xdx  du  sin xdx .
Thus
2) You must integrate the speed (not velocity) function to compute distance traveled . First compute
the vel)ocity function:
be 5. Thus
v(t )   a (t )dt   (2t  6)dt  t 2  6t  c .
v(t )  t 2  6t  5 .
Then the distance is

6
2
Since v(0)=5, the constant c must
t 2  6t  5 dt .
Making a sign graph,
t 2  6t  5 is
negative between 2 and 5, positive between 5 and 6. (first find the x-intercept:
x 2  6 x  5  0  x  1,5
Using the definition of absolute value
6
  (t 2  6t  5)dt  (
5
3) a)
n
n 

2
0
x 3 dx  lim
n 
 lim [
n 
n
i
i 1
B)
2
t3
t3
2
25
 25
34
 3t 2  5t ) 52 (  3t 2  5t ) 56  (  ( ))  (6  (
)) 
3
3
3
3
3
3
1
1  x2

2 sin 1 x
1  x2
ba 1
i
  xi  a  (x)i  1  .
n
n
n
2
b)
2

2
1
x 2 dx  lim
n 
n
 f ( x )x
i 1
i
i
 lim
n 
n
i 1
 f (1  n ) n
i 1

1
1
2
1
( )]  lim  [ 3 i 2  2 i  ]
n


n
n
n
i 1 n
2
n2

1
n
1 n(n  1)( 2n  1) 2 n(n  1) 1
 2
 n]
3
6
2
n
n
 i  n 1]  lim [ n
i 1
i 1
n 
ba 2
2i
  xi  a  (x)i  .
n
n
n
n
n
2i 2
2i 2
16
f ( )  lim  [( ) 3 ( )]  lim  [ 4 i 3 ]
n


n


n n
n
n
i 1
i 1 n
x0  a  0, xi 
n
n
 f ( x )x  lim 
i 1
2
5
t 2  6t  5 dt    (t 2  6t  5)dt
n
i 1
1
7
11 
3
3
6
x
i
1
n3

 f ' ( x)  e e e x
 [(1  n )
 lim [
=
ex
x0  a  1, xi 
n 
,
x2  1
1
1 2x
1
 f ( x)  (ln( x 2  1)  ln( x  1))  f ' ( x)  ( 2

) c)
x 1
2
2 x 1 x 1
f ( x)  e
 lim
x0
x x0
f ( x)  (sin 1 x) 2  f ' ( x)  2(sin 1 x)
f ( x)  ln
4) A)
x x
i
n 
i 1
16
16 n (n  1) 2
3
i
]

lim
[
 n n 4 4 ]  4
n 4 i 1
n
2
5) A) Recall that FTC part I says
d g ( x)
f (t )dt  f ( g ( x)) g ' ( x)  f (h( x)) h' ( x) .
dx h ( x )
1
1
1
1
dx  
b) Avg =
2
1
2
1 x
1  (sin x) 1  x 2
1
sin x
b


1
1
1
1 1
1
f ( x)dx   sin 2 xdx   [(  cos 2 x]) dx  C)

a
0
0
ba


2 2
2
ba
x0  a  0,  i 
 0.5  x1  0.5, x2  1, x3  1.5, x4  2 . Thus the midpoints are
4
a constant, g’(x)=0. similarly for h(x).
d
dx
In particular, if g(x) is

m1  0.25, m2  0.75, m3  1.25, m4  1.75 .
2
 (x
2
0
 1)dx  0.5[(0.25 2  1)  (0.75 2  1)  (1.25 2  1)  (1.75 2  1)]
6) A) First separate the denominator.
computed by the u-sub
function.

Thus


x 1
x
1
dx   2
dx  
dx .
2
x 1
x 1
1 x2
The first integral can be
u  x 2  1 , the second integral uses the formula for the inverse tangent
1
ln( x 2  1)  tan 1 x  c
2
b) let
u  tan 1 x  du 
dx
du
dx  
 ln u  c  ln tan 1 x  c
1
u
(1  x ) tan x
2
u  tan 1 x, dv  dx  du 
1
dx, v  x .
1 x2
 udv  uv   vdu  x tan
x
1
1
dx .
1 x2
Thus
c) Integration by parts:
Applying the parts formula,
x
1
 1
dx  ( x tan 1 x  ln( 1  x 2 )) 10   ln 2 (the second
2
2
4 2
1 x
integral can be computed using the u-sub)

d)

16
0
tan 4 xdx : u  4 x 
1


du  dx , x  0  u  0  0, x 
u  .
4
16
4


1 4
1
1
tan udu  ln sec u 04  ( 2  1)

4 0
4
4
solve for x:
e) (method 1) Let
Thus
1
u  3x  1  du  3dx  du  dx .
3
1
x  (u  1) .
3
1
1
( (u  1)) 2
3
1
2
2

x
1
1
u

2
u

1
1
3
2
du 
du 
1
 3x  1dx  3 
 (u  2u  u 2 )du
27 
27
u
u2
5
3
1
5
3
1
1 2 2 4 2
2
4
2

( u  u  2u 2 )  c 
(3x  1) 2  (3x  1) 2 
(3x  1) 2  c
27 5
3
135
81
27
2
Method 2)(not recommended) let
u 2  3 x  1  2udu  3dx  dx 
2
udu
3
u  3x  1 .
Square both sides to obtain
Now
Solve for x to get
1
x  (u 2  1) .
3

2
x
dx  
3x  1
1
( (u 2  1)) 2
2
3
( u )du 
u
3
5
3
1
2
2 u5 2 3
2
4
2
4
2
(3 x  1) 2  (3 x  1) 2 
(3 x  1) 2  c
(
u

2
u

1
)
du

(  u  u)  c 

135
81
27
27
27 5 3
7) A) Let
u  sec x  du  sec x tan xdx .
7
sec x
tan x sec xdx   7 u du 
1 u
1 sec x
7 c 
7
 c B) let
ln 7
ln 7
1
du
1
1
1
dx . 
dx   3   2  c  
c
3
x
x(ln x)
u
2u
2(ln x) 2
1
1
u  x  du 
dx  2du 
dx .
2 x
x
u  ln x  du 
\
1

x (1  x )
dx  2
1
du  2 ln 1  u  c  2 ln 1  x  c
1 u

the square root. Factor out 9 to make 1.
Let
1
9  4x 2
C) let
D) The inverse sine formula has 1 in
1
dx  
4
9(1  x 2 )
9
dx 
1
3
1
2
1  ( x) 2
3
dx .
2
2
3
x  du  dx  du  dx .
3
3
2
1
1 3
1
1
1
2
dx   
du  sin 1 u  c  sin 1 ( x)  c
2
3 2
2
2
3
2
1 u
1  ( x) 2
3
u
1
3
8) A) Integration by parts circular method: Let
 cos(ln x)dx  x cos(ln x)   sin(ln
x)dx .
1
u  cos(ln x), dv  dx  du   sin(ln x) dx, dv  x .
x
Apply parts again:
1
u  sin(ln x), dv  dx  du  cos(ln x) dx, dv  x .
x
have
 cos(ln x)dx  x cos(ln x)  x sin(ln
method),
Thus
 sin(ln
x)   cos(ln x)dx .
1
 cos(ln x)dx  2 [ x cos(ln x)  x sin(ln
u  ln( 2 x  5), dv  du  du 
Thus
2
dx, v  x .
2x  5
x)]  c

x)dx  x sin(ln x)   cos(ln x)dx .
Solving for
We
 cos(ln x)dx (this is the circle
B) Integration by parts:
ln( 2 x  5)dx  x ln( 2 x  5)  
2x
dx .
2x  5
To
integrate this, use the plus/minus trick:

).
2x
2x  5  5
5
5
dx  
dx   [1 
]dx  x  ln 2 x  5  c
2x  5
2x  5
2x  5
2
5
Therefore,  ln( 2 x  5) dx  x ln( 2 x  5)  x  ln 2 x  5  c .
2
(use u-sub to integrate
Alternatively, this problem can be done two steps: first u-sub of let
used in parts, we use k-sub. Then apply parts as follows:

1
dx
2x  5
u  2x  5 . However, since u is
1
1
dk  dx . Then  ln( 2 x  5)dx   ln k dk . Applying parts, let
2
2
1
u  ln k , du  dk  du  dk , u  k . Then
k
1
1
1
(uv   vdu)  (k ln k   (1)dk )  k ln k  k  c  (( 2 x  5) ln( 2 x  5)  (2 x  5))  c
2
2
2
k  2x  5 
Let
c) Use the double angle formula:
 sin
2
sin 2 3 x 
1
(1  cos 6 x) .
2
Thus
1
1
1
(1  cos 6 x)dx  x  sin 6 x  c

2
2
12
3xdx 
x2
x2  9  9 x2  9
9
9

 2
 2
 1 2
d) Use plus/minus trick:
.
2
2
x 9
x 9
x 9 x 9
x 9
x2
9
1
1 x
1 x
 x 2  9 dx   (1  x 2  9 )dx  x  9( 3 tan 3 )  c  x  3 tan 3  c
9) A)
1
1
1
sec u tan udu  sec u  c  sec 2 x  c .

2
2
2
1
1
Therefore, y  sec 2 x  (2 
)
2
2
y   sec 2 x tan 2 xdx  (u  2 x) 
2
1

1
.
sec( 2 )  c  c  2 
2
8
2
b) First find the base:
x 
20
 0. 5 .
4
A  0.5( f (0)  f (0.5)  f (1)  f (1.5))  0.5(0  0.52  12  1.52 )  1.75
dy
 xe x , y (0)  2
dx
Use the table method.
U
Dv
x
e 2 x
1

1 2 x
e
2
When
x

8
,y 2
We see from the graph below that the left end points need to
be used to obtain the lower sum.
c) Solve
Thus
0
1 2 x
e
4
 y
10) A) let
 (2
1  x 1 2 x
xe  e  c
2
4
(both times
1
1
1
u 10 du 
u 11  c 
(2 x  1)11  c

ln 2
11(ln 2)
11(ln 2)
e2x
(3 sin 2 x  2 cos 2 x)  c C) Integration by parts:
13
x
sin 1 xdx  x sin 1 x  
dx . This integral can
1 x2
1
1 x2

dx, v  x .
1

x
u  1  x 2 . Then  
x  0  u  tan 0  0, x 

8
 x  tan

4

1
2
dx 
1
1
tan 2 x sec 2 xdx   u 5 du  u 6
2
12

5
x
4  x4
 sec
3
dx  
x
4  ( x2 )2
dx 
du
4  u2
x tan x dx   sec 2 x(sec x tan x)dx :
2
2
 sec x sec x tan xdx   u du 
11) A)
1
2

1
2
E) First use u-sub before using the inverse sine formula. Since
f)
1

1
1
1
1
du   u 2 du  (2)u 2  1  x 2

2
2
2
u
1 x2
1
sin 1 xdx  x sin 1 x  1  x 2  c d) Let u  tan 2 x  du  sec 2 2 xdx .
2
be evaluated by u-sub

B) Using the circle method,
u  e 2 x , dv  cos 3xdx  e 2 x cos 3xdx 
u  sin 1 x, dv  dx  du 
Thus
1
9
1
1
9
 c  c  .  y   xe x  e 2 x 
4
4
2
4
4
u  2 x  1  du  (ln 2)2 x dx .
 1)10 2 x dx 
x
20
To find c, use x=0, y=2:
2
0
x 4  (x 2 ) 2 , let u  x 2 

.
1
96
1
du  xdx .
2
1 1 u
1
x2
sin
 c  sin 1
c
2
2
2
2
Now let
u  sec x  du  sec xdx .
u3
1
 c  sec 3 x  c
3
3
f ' ( x)  e 5 sin x (10 sin x)(cos x) B) f ' ( x) 
2
1 1
ln x x
c)
f ' ( x)  2 tan 1 x 
1
1 x2
d) take ln of both sides,
then differentiate both sides after bring down the exponent,
y  (ln x) x  ln y  x ln(ln x) 
12) Use
a (t )  32 .
v(t )  32t  32 .
First,
Next
1
1 1
1
y'  ln(ln x)  x
 y'  (ln x) x (ln(ln x) 
)
y
ln x x
ln x
v(t )   a (t )dt   (32)dt  32t  c .
Since
v(0)  32, c  32 .
s (t )   v(t )dt   (32t  32)dt  16t 2  32t  c .
ft, c=48. Thus s s(t )  16t
2
 32t  48 .
position is 0 at the ground level).
reach the ground. The speed is
Since the initial height is 48
Next find the time to hit the ground by solving
 16t  32t  48  0  t  2t  3  0  t  3 .
2
v(3)   32(3)  32  64
2
ft/s.
We have
s (t )  0
(the
It takes 3 second to
13) A) Let u=3x. du=3dx When x = 1, u = 3 and when x = 3, u=9.

3
1
1
1 9
1 1 1
dx   u 3 du  ( ) 2
3
3 3
3 2 u
(1  3x)
u 9
u 3
1 1 1
4
B) this function is not continuous at
 (  )
6 9 81 243
x = 0. Thus the integral cannot be computed in Math 180 !! C) Let
1
du  xdx . Solving for x 2 , x 2  u  1
2
3
2
x
x
1 u 1
1
1
1
1
2
2
 1  x 2 dx   1  x 2 xdx  2  u du  2  (1  u )du  2 (u  ln u )  c  2 (1  x )  ln( 1  x )  c
d g ( x)
14) A) Use FTC part I:
f (t )dt  f ( g ( x)) g ' ( x)  f (h( x)) h' ( x) . In particular, if g(x) is a constant,
dx h ( x )
d x2
1
1
g’(x)=0. similarly for h(x).
dt 
(2 x) b) use the inverse secant formula: Let
2

dx 1 (2  t )
(2  x 2 ) 2
u  1  x 2  du  2 xdx 
u  2x 

15) A)
b)

2
3
1
2

100
When

2 x (2 x) 2  4
2
4
,u 
.
3
3
x  1, u  2(1)  2, when x 
4
1 3
dx
1 1
u
 [ sec 1

2 2 u u2  4 2 2
2
u 0
u 1
1
 [sec 1
4

2
3
1
dx
2x 4x2  4
2
1 

 sec 1 1]  (  0) 
4 6
24
3
It is the area of a quarter of the circle of radius 2 centered at the origin
xdx  2
since it is a “signed area” of the triangle with base 2, height -2: ½(2(-2))=-2 c)
x 3 dx  0
: It is an odd function symmetric about the origin: thus positive portion and negative
2
100

dx
4  x 2 dx   :
0
0
1
du  dx .
2
portion cancels. d)

2
0
cos x dx  0 :
A trigonometric function over one cycle has zero signed area.
E) First switch the limits of integration and add negative. Then realize that this is a quarter of a
circle in the second quadrant. Thus

1`
0
1  x 2 dx   
0
1
1  x 2 dx  

4
1
u  1  sin 2 x  du  2 cos 2 xdx   du  cos 2 xdx . Thus
2
cos 2 x
1 1
1
1
 (1  sin 2 x) dx   2  u du   2 ln u  c   2 ln 1  sin 2 x  c B) First write
16) A) U-sub: let
and integrate by parts (twice). It is best to make a table(remember that
U
x
Dv
2
e 2 x
e
x2
 e 2 x dx
2 x
as
x
2
e  2 x dx
1
dx   e  2 x  c ) :
2
2x
1
 e 2 x
2
1 2 x
e
4
1
 e 2 x
8
2
0
e 2 x [
1 2 1
1
x  x  ] c
2
2
4
C) Integration by parts: It is recommended that you do not make a table since u is not a polynomial.
1
x3
x3
1
x3
1
2
. Thus  x ln x dx 
dx, v 
ln x   xdx 
ln x  x 2  c
x
3
3
3
3
6
1
1
1
dx   du  ln u  c  ln ln x  c
use u-sub: let u  ln x  du  dx Thus 
x
x ln x
u
sin1 x
1
1
e
1
dx . Thus 
e) U-sub: let u  sin x  du 
dx   e u du  e u  c  e sin x  c
1 x2
1 x2
u  ln x, dv  x 2 dx  du 
1 x
1
x
0 4  x 2 dx  0 4  x 2 dx  0 4  x 2 dx .
1
f) First separate the fraction as
1
D)
1
The first integral can be
computed using the inverse tangent formula. The second integral can be computed using u-sub:
1
1 x
1
x 1
1
1
5
dx  ( tan 1  ln( x 2  4)) 10  (tan 1  ln ) (the inverse tangent of ½ cannot be
2
2 2
2
2
4
0
simplified) g) U-sub: Method 1: let u  x  2  du  dx . When x = 0, u = 2 and when x = 1, u= 3.
 4 x
2
Solving for x gives
x  u 2.
1
Then
1
2
3
2
5
3
2
4
u 3
0 x x  2dx  2 (u  2) u du  2 [u  2u ]du  [ 5 u 2  3 u 2 ] u2
3
3
=I am too lazy to compute the value. But you must compute the value.
Method2: Let
u  x2.
u  x  2  2udu  dx . x  0  x  2 , x  1  u  3
1
3
3
2 5 4 3 3
2
4
2
0 x x  2dx   2 (u  2)(u)2udu   2 (2u  4u )du  ( 5 u  3 u ) 2 h) First observe that
e 2 x  (e x ) 2 . Use u-sub: let u  e x  du  e x dx . Then
Now square both sides to get
2
ex
1
1
1
x
 1  e 2 x dx   1  u 2 du  tan u  c  tan (e )  c
u  x  du 
 sin
1
2 x
dx  2 x du  dx  2udu  dx .
i) First u-sub, then parts (make a table). Let
Now
x dx  2 u sin udu  2u cos u  2 sin u  c  2 x cos x  2 sin x  c
Second method: square both sides:
u  x  u 2  x  dx  2udu  dx .
 sin
Then
x dx  2 u sin udu  2u cos u  2 sin u  c  2 x cos x  2 sin x  c
i_) First u-sub. Then separate the exponent as
u  x 2  du  2 xdx 
1
du  xdx .
2
Then

x3  x 2  x .
Finally apply parts (make a table) . Let
x 3e x dx   x 2 e x xdx 
2
2
1
1
ue u du  (ue u  e u )  c

2
2
2
1 2 x2
(x e  ex )  c
2
1
2
j) Let u  x  du  xdx . Thus
2
1
1
1
2
2
2
 x sec x dx  2  sec udu  2 ln sec u  tan u  2 ln sec x  tan x  c
ba 2
2i
17) A) x 0  a  2, xi 
  xi  a  (x)i  1  .
n
n
n
n
n
3
2
i
2
2
f ( xi )xi  lim  f (1  )

1 ( x  2 x)dx  lim
n 
n 
n n
i 1
i 1
n
n
2i
2i 2
8
16
6
 lim  [((1  ) 2  2(1  ))( )]  lim  [ 3 i 2  2 i  ]
n 
n 
n
n n
n
n
i 1
i 1 n
n

n
8
16
6
8 n(n  1)( 2n  1) 16 n(n  1) 6
 lim [ 3  i 2  2  i  1]  lim [ 3
 2
 n]
n  n
n  n
n i 1
6
2
n
n i 1
n
i 1
16 16
50

 6
6
2
3

n
b)

lim
n
1
0
i
n
i 1
xdx 
2
1 n i
.

n n
i 1 n
 lim
You must recognize that this is a Riemann sum o
1
2
18) A) use FTC part I
an upper sum.
2 x sec 4 ( x 2 )
b) First find the base:

20
 1.
2

1
0
xdx .
Thus the sum is
Use the left end points to obtain
A  1( f (0)  f (1))  1((4  02 )  (4  12 ))  7
n
19) You must recognize that
i
lim  ( x)
n 
n
i 1
3
is the Riemann sum for
1
x
0
1
3
dx .
Thus the limit is ¾
20)
Using dx, you need to two integrals since the top equation changes at x = 0.
0
1
1
0
A   ( x  1)dx   (1  x)dx
the same.
b) You just need one integral since the left and the right equation stay
1
A   [(1  y)  ( y  1)]dy
0
21) A) Method 1: First u-sub: you need to solve for x:

x
2
x 1
dx  
(u  1)
5
2
2
u
3
2
1
2
du   [u  2u  u
3
1

2
u  x  1  du  dx, x  u  1  x 2  (u  1) 2 .
]du 
5
2
3
2
Thus
1
2
2
1
u  u  2u  c
5
3
1
2
1
( x  1)  ( x  1) 2  2( x  1) 2  c
5
3
2
2
Method 2) : let u  x  1  u  x  1  x  u  1  dx  2udu .
u4
x2
(u 2  1) 2
u 4  2u 2  1
1
3

 u 2  ln u  c
dx

du

du

(
u

2
u

)
du
 x 1  u


4
u
u
1
6
3 2
3
2
B) First write x  (x ) and let u  x  du  x dx . When x= 0, u= 0, x= 1, u= 1. Then
3
1
2
x
1 1 du 1
1

1
1
1
1
0 1  x 6 dx  3 0 u 2  3 tan u 0  3 (tan 1  tan 0)  12

c) First observe that
cos3 2 x  cos 2 3x(cos 3x)  (1  sin 2 3x)(cos x) .
1
u  sin 3x  du  cos 3x(3)dx  dx  cos 3xdx .
3
 cos
3
Use u-sub:
1
(1  u 2 )du

3
2 xdx   (1  sin 2 3x)(cos 3x)dx 
1
u3
1
1
 (u  )  c  (sin 3x  sin 3 3x)  c
3
3
3
3
2
2i 

lim  ( 3   x)
n 
n
i 1 
n
22) A) You must recognize that
xi  0 
23) Since
is the Riemann sum for
2
i Thus the limit is 98/3 by integrating the function.
n
a(t )  3, v(t )   v(t )dt   3t  c .
distance traveled, we compute

4
1
B)
 (3  x
0
2
)dx
since
x2
2x
2
 lim x  lim x  0
x  e
x e
x  e x
lim
Since the initial velocity is 6,
 3t  6 dx .
2
Using the sign graph,
v(t )  3t  6 .
 3t  6  0
To find the
between t=1 and 2.
2
4
3 2
3 2
9
15
2

3
t

6
dx

(

3
t

6
)
dt

1
1
2  (3t  6)dt  ( 2 t  6t ) 1 ( 2 t  6t )  (6  2 )  (6  0)  2
d g ( x)
24) Recall that FTC part I says
f (t )dt  f ( g ( x)) g ' ( x)  f (h( x)) h' ( x) . In particular, if g(x) is a
dx h ( x )
4
Thus
 
3
d
ln t dt   ln x 3 (3x 2 )

dx x3
constant, g’(x)=0,. similarly for h(x).
b)
1
2x
(1  x 2 ) 2
c)
2 x tan x 2  4 x 3 tan x 4
25)
a) First write the exponent as

4
x 1
1
c)
4
dx   (
1
x

3
0
x
x  2 dx :
x

x8  (x 4 ) 2 .
1

Let
1
u  x 4  du  4 x 3 dx  4
0
3
1

du  tan 1 u 10 
2
1 u
4
b)
1

4
2
8
)dx   ( x 2  x 2 )dx  ( x 2  2 x 2 ) 14 
1
3
3
x
1
The graph is broken up into two pieces: the left portion of V and the right portion of V. Thus
you need to separate the integral at x = 2.

3
0
2
3
0
2
x  2 dx    ( x  2)dx   ( x  2)dx 
5
2
d) The integrand is odd, integrated over a symmetric period about 0. Thus it is 0 (an odd function divided
by an even function is an odd function.)
26)
dP
dP
dP
 kP is separable:
 kP 
 kdt  ln P  kt  c  P  Ae kt . When t = 5, P=2A . Thus
dt
dt
P
ln 2
2 A  Ae
27) a)
ln 2
k 
. Let P= 3A and solve for t: 3 A  Ae
5
5k
e
e
e 2 x  (e x ) 2  (u  1) 2 Then

b)
ln 2

5
t  ln 3  t 
3
2
1
2
7
3x
1  e x dx   e 2 x 1  e x e x dx   (u  1) 2 u du =
5
3
7
5
3
2
4
2
2
4
2
(u  2u  u )du  u 2  u 2  u 2  c  (1  e x ) 2  (1  e x ) 2  (1  e x ) 2  c
7
5
3
7
5
3
1
 x(log
3
1
 x(log
3
x)
2
5 ln 3
ln 2
1  e x dx : First observe that e 3 x  e 2 x e x . let u  1  e x  du  e x dx and e x  u  1 and
3x
5
2
t
5
x) 2
dx . Change of base formula first, then u-sub.
dx  
u  ln x  du 
1
1
dx  (ln 3) 2 
dx Let
2
ln x 2
x
(ln
x
)
x(
)
ln 3
1
1
1
1
(ln 3) 2
2
2
dx  (ln 3)3 
dx

(ln
3
)
du

(ln
3
)
(

)

c


c
 u2
x
x(ln x) 2
u
ln x
c)

ln x
1
1
1
dx : apply parts: u  ln x, dv  2 dx  du  dx, v   Using the parts formula,
2
x
x
x
x
uv   vdu  

d)
ln x
1
ln x 1`
  2 dx  
 c
x
x
x
x
xe x dx : Let u  x  du 
1
2 x

dx  dx  2 x du  2u du .
x e x dx   ue u 2udu  2  u 2 e u du
Then the problem can be completed by applying parts twice(you can make a table:
U
Dv
u2
eu
2u
eu
2
eu
0
eu
2 u 2 e u du  2(u 2 e u  2ue u  2e u )  c  2( xe
x
 2 xe
x
 2e
x
)c
e) The integral cannot be computed in Math 180 since the function is not continuous at x=-1
and x=1..
28)
a)
Use FTC Part II
f ( x)  
1
3
sin1 x
x 1
2
1
dx  f ' ( x)  
(sin
1
1
x)  1 1  x 2
2
b) This is a separable DE.
First factor the right side using the grouping method (Math 51 technique) . Then separate x and y and integrate both
sides.
dy
dy
1
 x y  x  y 1 
 ( y  1)( x  1)  
dy   ( x  1)dx
dx
dx
y 1
2
3
 ln y  1 
2
use the inverse secant formula: Let
u  4x 
1
du  dx .
4
dx
  4 34
b)
3
x   , u  6.
2
3
3
4
3

2


x 16x 2  9
dx
29) a)

2

dx
1
u
 [ sec 1
6
3
u u2  9 3

2
x 2  x c
x2  x
x2  x
2 2
x  x  c  y 1  e3
 y  1  e c e 3
 y  ce 3
1
3
x 3e x dx
u 3
u 6
3
  34

2
x ( 4 x) 2  9


3
2
1
1

 [sec 1 1  sec 1 2]  (1  )
3
3
3
When
3
3
x   , u  4( )  3, when
4
4
dx
4 x (4 x) 2  9

Let u  x  du  2 xdx .
2
x e
3
x2
dx   x  x 2  e x dx 
2
by applying parts once (make a table)
2 tan
b)


2 tan
c)

x
x
1
ue u du . Then the problem can be completed
2
2
2
1
1
1
ue u du  (ue u  e u )  c  ( x 2 e x  e x )  c

2
2
2
sec 2 x
sec 2 x
dx Use u-sub: u  tan x  du 
dx 
x
2 x
sec 2 x
x
dx  2 2 u du  2
2u
2 tan x
c  2
c
ln 2
ln 2
tan 2 4 xdx   (sec 2 4 x  1)dx 
1
tan 4 x  x  c
4
30)
a) First use the double angle identity. Then separate the integral. Use parts for the second integral.

1
1
1
1 x2 1 1
1
x cos 2 xdx   x( (1  cos 2 x)dx   xdx   x cos 2 xdx 
 ( sin 2 x  cos 2 x)  c
4
2
2
2
2 2 2 2
U
Dv
X
cos 2x
1
1
sin 2 x
2
0
1
 cos 2 x
4
b) Use inverse tangent formula.
cos x
 1  sin
2
x
dx  
u  sin x  du  cos xdx ,
du
 tan 1 u  c  tan 1 (sin x)  c
2
1 u
c) This is a trick question:

3
3
x5  x
ex
2
dx  0
since
x5  x
ex
2
is an odd function as an odd function divided
by an even function is an odd function.. When an odd function is integrated from –a to a, its signed
area is 0.
31) Since
a (t )  9.8, v(t )    9.8dt  9.8t  c .
Since the initial velocity is 29.4,
s (t )   v(t )dt   (9.8t  29.4)dt  4.9t 2  29.4t  c .
level. Thus
v(t )  9.8t  29.4 .
Thus
But c=0 since the projectile id fired from the ground
s(t )  4.9t 2  29.4
a) To hit the speed of 9.8 m/sec, the velocity can be 9.8 or -9.8.
the way up) and
v(t )  9.8  9.8t  29.4  9.8  t  2
(on
v(t )  9.8  9.8t  29.4  9.8  t  4 (on the way down).
b) To find the maximum height, first set
v(t )  0  9.8t  29.4  0  t  3 . s(3)  4.9(3) 2  29.4(3)  44.1m