Sample Final Solutions:
1)
x2  9
 x dx : Trig sub: Let x  3sec . Thus
 dx  3 sec  tan d , x 2  9  (3 sec  ) 2  9  9 tan 2   3 tan 
a)
x2  9
3 tan  (3 sec  tan  )
dx  
d  3 (tan 2  )d  3 (sec 2   1)d  3[tan    ]  c
x
3 sec 

bnbbNow make a triangle using
 3[sec  tan    ]  c 
x2  9
x
 sec 1  c
3
3
1
1
x dx   ( (1  cos 2 x)) 2 dx   1  2 cos 2 x  cos 2 2 x)dx 
2
4
1 1
3
1
1
(1  2 cos2 x   cos 4 x)dx  x  sin 2 x 
sin 4 x  c
2 2
8
4
32
b) Use double angle twice:
1
4
x
 sec 
3
 sin
4
7x2  x  9
dx :
x3  x
7x 2  x  9
 2x  1 9
dx   ( 2
 )dx .
c) 
Partial fractions will give (see below for details) 
3
x x
x 1 x
7x2  x  9
dx   ln( x 2  1)  tan 1 x  9 ln x  c
Separate the first fraction and use u-sub for the first part. 
x3  x
7 x 2  x  9 Ax  B C
 2
  7 x 2  x  9  x( Ax  B)  C ( x 2  1) . First
(For the partial fractions,
2
x( x  1)
x 1 x
let x=0. We get C = 9.
7 x 2  x  9  x( Ax  B)  9( x 2  1)  ( A  9) x 2  Bx  9 .
By equating
coefficients, we get A = -2, B=1.
u  (ln x) 2 dv  dx  du 
d) Use parts twice:
x(ln x) 2  2 ln x dx :
 ln xdx  x ln x  
Use parts for
 ln xdx :
2 ln x
dx, v  x :
x
apply the parts formula to obtain
u  ln x, dv  dx  du 
1
xdx  x ln x   1dx  x ln x  x .
x
Therefore,
1
dx, v  x .
x
 (ln x)
2
dx 
x(ln x) 2  2 ln x dx  x(ln x)2  2( x ln x  x)  c .
a) e)
e
3x
cos 4 xdx :
the integral.
Use the circle method: Use parts twice. We can make a table to summarize
U
Dv
e
3e 3 x
cos 4x
sin 4 x
4
9e 3 x
Ax
3x


f)
e3 x sin 4 x 3 3 x
9
 e cos 4 x   e3 x cos 4 xdx .
4
16
16
16
1
3
e 3 x cos 4 xdx 
( e 3 x sin 4 x  e 3 x cos 4 x)  c
25 4
16
e3 x cos 4 xdx 
 sec
3
Solving for
e
3x
cos 4 xdx ,
x tan 3 x dx   sec 2 x tan 2 x(sec x tan x)dx   sec 2 x(sec 2 x  1) sec x tan xdx .
Let
u  sec x  du  sec x tan xdx .
u5 u3
(sec 5 x) (sec 3 x)

c 

c
5
3
5
3
1
1
dx  
dx : Complete the square first, then use trig sub: 
x 2  4x  5
x 2  4x  5

sec 2 x(sec 2 x  1) sec x tan xdx   u 2 (u 2  1)du 
g)

1
( x  2) 2  1
dx .
x  2  tan   dx  sec 2  d . We have ( x  2) 2  1  tan 2   1  sec  . Therefore
1
1
2
2
 x 2  4x  5 dx   sec sec  d   sec d  ln sec  tan   c  ln x  4x  5  ( x  2)  c
1
1
1
2
2
2
h) Use two double angle formulas: sin x cos x  sin 2 x to write (sin x cos x )  ( sin 2 x )  sin 2 x and
2
2
4
1
sin 2 2 x  (1  cos 2 x) We have
2
1
1
4
2
2
2
2
2
 sin xdx cos xdx   sin x cos x sin xdx  m : double angle  4  sin 2 x( 2 (1  cos 2 x))dx =
1
1
1
1
sin 2 2 xdx   sin 2 2 x cos 2 xdx   (1  cos 4 x)dx   u 2 du (for the second integral, let

8
8
16
16
1
1
1
1
u  sin 2 x  du  cos 2 xdx )= 
x  sin 4 x  sin 3 2 x  c
2
16
64
48
x
1
dx : first u-sub. Write 1  x 4 as 1  ( x 2 ) 2 and let u  x 2  du  xdx .
i) 
4
2
1 x
x
1
1
1 1
1 1 2
 1  x 4 dx  2  1  u 2 du  2 sin u  c  2 sin x  c
1
3
2
2
3 x2
3 x2
2 x
u
j)  x e dx : u  x  du  2 xdx  du  xdx . Separate x  x  x ,  x e dx   x e xdx   ue du 
2
Next let
u
U
Dv
1
eu
0
eu
= ue
eu
u
 eu  c  x 2e x  e x  c
2
2
2) First observe that they are sequences, not series.
a) Multiply the top and bottom by the conjugate:
n  1  n  lim
n 1  n
n 1  n
1
0
n 1  n
n 1  n
n!
n n 1 n  2 2 1
1
n n 1 n  2 2
b) 0  lim n  lim


   lim  0 ( the expression 

 can be dropped
n n
n n
n
n
n n n n
n n
n
n
n!
1
n!
n!
since it is less than 1) . Using the squeeze theorem, 0  lim n  lim
 0  lim n  0  lim n  0
n  n
n  n
n  n
n  n
1 sin n 1
1
1
c) We know that  1  sin n  1 . Thus  
 and lim ( )  0, lim  0 . Thus by the squeeze
n 
n  n
n
n
n
n
sin n
theorem, lim
0
n 
n
lim
n 
n 
 lim
n 
3)
a) Use the ratio test: the ratio is
1
, thus convergent
3

(n 2  1) n
1 


1  2 


2n
n
n 
n 0
n 0 

b) The root test fails.
Next try TOD.
n
Apply the test of divergence using
1 n
) , take ln of both sides and bring down the exponent to get
n2
1
2
( 3 )
1
1
n
ln( 1  2 )
1 2
1 n
1
n  lim
n
lim ln(( 1  2 ))  lim
 lim
 0 . Since ln y  0 , y converges to
n
n
n
n
1
1
n
 2
n(1  2 )
 2 
n
n
 n 
L’hopistal” : To compute lim (1 
n 
e0  1.

c)
Thus the series is divergent by the test of divergence.
1
 (ln 2)
n 2
n
: Use geometric series test:
r
1
 1.
ln 2
d)Convergent by the comparison test: recall that
Thus the series is divergent.
ln n  n c for any power of c. Pick any c that makes the
1
exponent of n greater than 1.
ln n

n2 1
n2
2
n
n 
2
2
2
1
n
3
2
1
and
n

e) Use the comparison test: guess is convergent since
3
2
is convergent by the p-test.
n 1
3
n
n
n 1
behaves like n and p=2. Thus you
construct a larger p-series by making the top bigger, the bottom smaller.
the series is convergent by the comparison test.
f) Use the integral test. Let
1
u  ln x  du  dx .
x
divergent by the integral test.
Then


2
1
x ln x
dx  lim
n 1
nn
1

4 2
3
3
n 1
n
n
n3 
2
b
u
b  ln 2

1
2
du  lim 2 u
b 
b
ln 2
. Thus
 ,
g) To use the comparison test, you must first take the absolute value since the test requires the terms to
cos 4n
be positive. Then use the theorem that states absolute convergence is convergence.

1

2
n 2 n
h)

is convergent by the p-test. Thus
cos 4n
2
n
n
n2
is convergent. Therefore

1
n2
and
n n

cos 4n
is convergent.

2
n2 n  n
2

2n
2 n
is
convergent
by
the
geometric
series
test.
Therefore
(
)

 n n
n 03 5
n  03 3  5

2n
2n
2

 ( )n
n
n
n
3 5
5
5
and
is
convergent by the comparison test.
i) Apply the ration test:
a n 1
1  3  5 (2n  1)( 2n  1) 1  3  5 (2n  1)


 lim
n 1  4  7  (3n  2)(3n  1)
n  a
1  4  7  (3n  2)
n
lim
2n  1 2
1  3  5(2n  1)( 2n  1)
1  4  7 (3n  2)
  1.

 lim
n 
1  3  5(2n  1)
1  4  7 (3n  2)(3n  1) n 3n  1 3
lim
Thus the series is convergent
by the ratio test.

j) Since ln(n) grows very slowly,
1
 (ln n)
n 2
2
would diverge. Use the fact that
ln n  n any power
for large n. .
So use any power less than 0.5 (using a power more than 0.5 would make the series too small).
ln n  n 
1
1
1


2
2
(ln n)
n
( n)

and
1

n 2 n

is divergent by the p-test. Thus
1
 (ln n)
n 2
2
is divergent by the
comparison test.
k)
 cos 4 x cos 6 xdx :
Use the product to sum formula: Since
cos( A  B)  cos A cos B  sin A sin B, cos( A  B)  cos As cos B  sin A sin B ,
1
1
cos A cos B  [cos( A  B)  cos( A  B)]  cos 4 x cos 6 x  [cos 2 x  cos 10 x] . Thus
2
2
1
1 1
1
 cos 4 x cos 6 xdx  2  (cos 2 x  cos 10 x)dx  2 [ 2 sin 2 x  10 sin 10 x]  c
4)
tan 1 x , then replace x by 3x 2 .



(1) n x 2 n1
2 n
n 2n
( x ) dx    (1) x dx  
c

2n  1
n 0
n 0
n 0
a) First find the power series expansion of
tan 1 x  
1
dx  
1 x2
We have
By evaluating the expression at x

(1) n (3x 2 ) 2 n1
(1) n 32 n1 4 n2

x
2n  1
2n  1
n 0
n 0
b) First find the power series expansion for ln( 1  x) . Then use the box principle to replace x

1
1
  (1) n x n and ln( 1  x) can be obtained by integrating
that
.
1 x
1  x n 0

= 0, we get c = 0. Thus
ln( 1  x)  
1
dx  
1 x
tan 1 (3x 2 )  

n 0

Let x = 0 :
ln( 1  x)   (1) n

ln( 1  x 2 )   (1) n
n 0

 (1) n x n dx    x n dx    (1) n
n 0
2n 2
x
n 1
n 0
n 0
x n1
 c  ln1  c  c  0
n 1
x n1
c.
n 1
x2.
Next find c:

Thus
by
ln( 1  x)   (1) n
n 0
x n1
.
n 1
Therefore,
Recall
c) First observe that
d2
1
2
1
1 d2
1
(
)



(
).
2
3
3
2
dx 1  x
(1  x)
(1  x ) 2 dx 1  x
1
1 d2
1
1 d2

(
)
(1  x)3 2 dx 2 1  x
2 dx 2
1 
  n(n  1) x n1
2 n 0
1 
x   n(n  1) x n2 .

2 n 0
n 0
Thus

n
Therefore,
x3
1 3 

x  n(n  1) x n2 
3
2 n 0
(1  x)
5)
a) This is separable. First separate x and y. Then integrate both sides.
dy e 2 x  y

  ye  y dy   e 2 x dx .
dx
y
U
Y
Integrate the left hand side using parts:
Dv
ey
 e y
ey
1
0
  ye  y  e  y 
1 2x
e  c (this is an implicit solution)
2
dy
 xy 2  y 2  x  1 separable: first fact xy 2  y 2  x  1  ( y 2  1)( x  1)  ( y  1)( y  1)( x  1) .
dx
dy
dy
 xy 2  y 2  x  1  
 ( x  1)dx . But using partial fractions,
dx
( y  1)( y  1) 
dy
dy
1
1 1
1
 [

] . Thus
 xy 2  y 2  x  1  
 ( x  1)dx =
( y  1)( y  1) 2 y  1 y  1
dx
( y  1)( y  1) 
1
1
1
1
x2
(

)dy   ( x  1)dx  ln ( y  1)( y  1) 
c
2  y 1 y 1
2
2
b)
Thus
6)
a) First find the points of intersection by solving
3 sin   1  sin    
find the area in the first quadrant: then double it. From
crosses only
3sin  .
From

Therefore, area =

6
to
0 to
 5
6
,
6
. a)Use symmetry:

, a typical line through the origin
6

, a typical line through the origin crosses only 1 sin  .
2

1
2  [  6 (3 sin  ) 2 d  2 (1  sin  ) 2 d ]
2 0
6
b) A typical line through the origin crosses enter the region through
3sin 
. Area =
c) First set r = 0:
1 sin  an
d exits through
5
6
1
2
2
 [(3 sin  )  (1  sin  ) ]d
2 6
1  2 sin   0   
7 11
,
6 6
11
.
1 6
2
7 [((1  2 sin  ) ]d

2 6
7)
a) Need to find the length and the depth of the typical horizontal strip. Place the diameter of the sphere
on the x-axis. Take a typical horizontal piece . Its length is
is
  (length)(de pth) y    (2 x)(9  y)y .
circle of radius 5,
x 2  y 2  25  x  25  y 2
(no need to use
  (2 x)(5  y)y    (2 25  y 2 )(9  y)y .
to 5,
   2 25  y 2 (9  y)dy
5
Thus its Riemann sum
Next express x as a function of y. Since (x,y) lie on the
Thus
5
2x, its depth is 9  y .
As
 since x is place on the positive side).
y  0 , since the typical piece moves from -5
b) Place the bottom of the triangle on the x-axis. Take the typical horizontal piece. Its length is
Using the famous
Riemann sum is
2  1  3 triangle, we know the height of the triangle is 2 3 .
  (length)(de pth) y    (2 x)((3  2 3)  y)y .
2x,
Thus the
Next express x as a
function of y. Find an equation of the line through
(0,0) and (2, 2 3) . Using Math 50, an equation
y
y
is y  3 x  x 
Thus   (2 x)((3  2 3  y )y    (2
)(3  2 3  y)y . As y  0 ,
3
3
2 3 2
since the typical piece moves from 0 to 2 3 ,   
y(3  2 3  y)dy
0
3
8)
Washer : use dx since you pick a typical piece perpendicular to the rotating axis. In washer you need to
find the outer radius and inner radius.
1
  [(2  ( x)) 2  (2  x) 2 ]dx
0
Shell:
In shell, you need to find a) shell radius b) shell height
use dy since a typical piece is taken parallel to the rotating axis. You need to break up the region into
two sections since the left equation changes at y = 0. Also since you are using dy, you must solve the
equations for x.
1
0
0
1
2 [ (2  y)(1  y)dy   [(2  y)(1  ( y))dy]
9)
a)
mmn
(x 2 )   62.4x 2 y .
The weight (force) of the typical piece is (volume)(density) =
to express x as a function of y:
weight is
As
Use similar triangles
x 9
y

 x  . (or an equation through (0,0) and (9,18) Thus the
y 18
2
y2
(62.4)( )( )y  15.6y 2 y .
4
The distance is 18-y. Thus Work
 15.6)( )( y 2 )(18  y)y.
8
y  0 W  15.6  y 2 (18  y)dy
0
(x 2 )   62.4x 2 y .
(62.4)( )(100  y 2 )y .
c) The weight (force) of the typical piece is (volume)(density) =
equation
x  y  100  x  100  y
2
2
is 5-y. Thus Work
2
2
Thus the weight is
  (62.4)( )(100  y 2 )(5  y)y.
d) First find the spring constant k:
As
F  kx  3  2k  k 
Use the circle
The distance
10
y  0 W  62.4  (100  y 2 )(5  y)dy
10
3
.
2
Then Work =

b
a
kxdx 
3 4
xdx  24 J
2 0
10)
a) Apply the ratio test:
lim
n 
a n 1
an
(3x  1) n 1
(3x  1) n1
2n 2  1
2(n  1) 2  1
 lim
 lim

n 
n  (3 x  1) n
(3x  1) n
2(n  1) 2  1
2n 2  1
2n 2  1
2n 2  1

(
3
x

1
)
lim
 3x  1 .
n 2n 2  4n  3
n 2n 2  4n  3
( 3x  1 ) lim
2
3x  1  1  1  3x  1  1    x  0 .
3
Now set this less than 1 and solve for x:

1
(3x  1) n



2
2
n 1 2n  1
n 1 2n  1

Now test the end points: first x=0.
. This is convergent using the comparison test:
1
2n  1
2

1
2n 2
and
1
2n 2
is convergent by the p-test.

(3x  1) n
(1) n
.



2
2
n 1 2n  1
n 1 2n  1

Next at x=-2/3:

 2n
n 1
1
2
1
is convergent. Thus the interval of convergent is
b) Apply the ratio test:
lim
n 
a n 1
an
(2 x  5) n

n!
n 1
f ( x) 
2
[ ,0]
3
(2 x  1) n 1
1
(2 x  1) n1 (n)!
(n  1)!
 0  1.
 lim
 lim
 2 x  5 lim
n  n  1
n  ( 2 x  5) n
n  ( 2 x  5) n ( n  1)!
n!

Thus the radius of
11) )
This is convergent absolutely since we just showed that
is
 , the interval is (, )
1
2
8
48
 f ' ( x)  
, f " ( x) 
, f ( 3) ( x )  
2
3
1  2x
(1  2 x)
(1  2 x)
(1  2 x) 4
1
2
8
. Thus the second degree Taylor polynomial is
f (2)  , f ' (2)   , f " (2) 
5
25
125
1 1
4
 ( x  2) 
( x  2) 2 . For the error, the maximum f ( 3) ( x ) for 1.9  x  2.2 is attained when
5 25
125
48
x=1.9. M 
. Also, the maximum distance from a=2 for 1.9  x  2.2 is 0.2. Thus
4 .8 4
48
4
error  4.8 (0.2) 3
3!
Thus at a=2,
12)

a) Memorize
3n x n 2

n!
n 0
xn
e 
n 0 n!
x

, radius is
. Substitute 3x into x, then multiply by
x
2

f ( x)  x e
2 3x
(3x) n
x 

n!
n 0
2

b) Use the double angle formula.
f ( x)  sin 2 2 x 
(1) n x 2 n
 (2n)! . Thus
n  0
1  (1) n (2) 2 n1 ( x) 2 n
(the first term
 
2 n1
(2n)!
1
(1  cos 2 x) .
2
Use
cos x 

1 1  (1) n (2 x) 2 n 1 1  (1) n (2) 2 n ( x) 2 n
1 1
 cos 2 x   
  
2 2
2 2 n0 (2n)!
2 2 n0
(2n)!
1
1
1
of the summation is  . Thus the term cancels with , so
cancels. The radius is  .
2
2
2
1
1
1
1
x2 2

 (1  ) . Use the binomial formula
c) Observe that f ( x) 
9
9  x2
 x2  3
9 ( 1   
 9 
1
1

 
k  n 1
x2 2
   x 2n x 2
k
(1  x)    x .
(1  )    2  n ,
 1  x  3 so radius is 3.
3
9
9
9

n 0 
n 0  n 
n 
13)
a)
x  2 sin t , y  2 cos t ,0  t  2
d)
Eliminate t using the identity
a) Range:
sec 2 t  tan 2 t  1  x 2  y 2  1 . This is a hyperbola crossing the x-axis.




x  sec t ,  t   x  1 . y   tan t   t     y   , y takes on all real
2
2
2
2
numbers.
Direction:
c)
dx
 sec t tan t
dt
x   t , y  t 2  1.
A) Range:
Since
:
dy
dx
can be positive or negative.
  sec 2 t  0 .
dt
dt
x   t ,  t   x  t  ( x) 2  x 2 .
x   t , so x is never
y  x4 1
positive.
Thus
Thus y is decreasing.
y  t 2  1  ( x 2 ) 2  1  x 4  1.
y  t 2  1  y  1 , Thus the graph is only the left side of
I
B) direction: since
dx
1
dy
dy

 0 , it is traced to the left.
 2t , t  0 
0.
dt
dt
dt
2 t
increasing.
14) ) a) This is an improper integral:


1
b
x
x
1
1
dx  lim  2 dx  lim ( ln( x 2  1) 1b  [lim ln( b 2  1)]   , divergent.
b  1 x  1
b  2
2 b 
x 1
2
: divergent
b) The function is undefined at x=1. Approach 1 from the right hand side.
lim 
b 1
15)
1
dx  lim [ln 2  ln( b  1)]   :
b x 1
b 1
3
divergent.
Thus y is always
x2 x3
x3 x4

 ..)  x  x 2 

 ...
2! 3!
2
6
0.1
0.1
x3 x4
x2 x3 x4 x5
0.12 0.13 0.14
x
2
0.1
xe
dx

(
x

x



...)
dx

(




...)

(


 ) .
0
0
0
2
6
2
3
8 30
2
2
8
0.1
0 . 14
0.12 0.13
4
xe x dx 

By trial and error,
is the first term with value less than 10 . Thus 
0
8
2
3
k
(
k

1
)
k
(
k

1
)(
k

2
)
k
Use the binomial formula: (1  x)  1  kx 
x2 
x3  .
2
6
1 1
1 1
3
1
1
1
( ) 2
( )(  ) 2
2 1
2
x
x
x
2 ( x )3  
(2  x 2 ) 3  2 3 (1  ) 3  2 3 [1 
 2 2 ( )2  2 2
2
2
2
2
6
2
1
1
1
2
2
0.14
x
1 4
1 6
0.1
1
1
 2 3 [1 

x 
x  00.1  2 3 [1 
 0.14 
0.16   By trial and error, 2 3
is
32
2 32
128
2
32
128
a) )First observe that the Maclaurin series for
b)
the first term with value less than
10
4
xe x
is
x(1  x 
. Thus the integral

0.1
3
0
1
3
2  x dx  2 [1 
2
0.12
]
2

16) First check for the absolute convergence: use the comparison test to show
 2n
n 1
n
2n  1
2

n
11

.
2
3n
2n  n
2
n
2
1
is divergent.
The smaller series is divergent by the p-test. Thus the series does not
converge absolutely.
Next check for conditional using the alternate series test.
a)
b)
d
n
 2n 2  1
(
)
The terms are decreasing since its derivative
is negative.
dn 2n 2  1 (2n 2  1) 2
n
 0 . Thus the series converges conditionally.
Using L’hopital, lim
2
n  2n  1
17 ) Use Newton’s law of cooling: We can set 12 pm to be t=0. Then at 1 pm t=1. We have
T (0)  80, T (1)  75 . We need to find t when T=98. The room temperature TM  70
dT
dT
 k (T  70) 
 kdt  ln T  70  kt  c  T  70  e c e tt . Now remove the absolute value
dt
(T  70)
c
 kt
by taking  on the right hand side. We let  e  A . We have T  70  Ae . When t=0, T=80.
1
1
80  70  Ae0  A  10 . When t=1, T=75: 75  70  10e k  e k   k  ln( )   ln 2  k  ln 2
2
2
 t ln 2
The equation now becomes T  70  10e
. Find t when T=98.
ln
2
.
8
98  70  10e t ln 2  2.8  e t ln 2  t  
 1.5 . Thus the murder took place at 10:30 a.m.
ln 2
18)
a) This is a geometric series with common ratio
1
e
and
a1  1 . Thus

e
n 0
n

a1

1 r
1
1
1
e

e
e 1
(1) n ( ) 2 n1
 sin   1

(2n  1)!
n 0
2
1
1
c) This is telescoping. First apply the partial fraction technique to get
. Then


(2n  1)( 2n  1) 2n  1 2n  1
1 1
1 1
1 1
1
1
1
th nth partial sum is s n  (  ) )(  )  (  )    (
Thus

)  1
1 3
3 5
5 7
2n  1 2n  1
2n  1

2
1
 lim (1 
) 1

n 
2n  1
n 1 ( 2n  1)( 2n  1)
b) This is the power series of sinx with
19)a) Use the formula
L
b
a
x   . Thus
dx
dy
( ) 2  ( ) 2 dt .
dt
dt

4
0

1
4
3
9
1  ( t 2 ) 2 dt   1  t dt :
0
2
4
1
let
3
4 10
4 2
9
9
u  1  t  du  dt , t  0  u  1, t  4  u  10 . Thus  u 2 du  ( (10 2  1))
1
4
4
9
9 3
2
3
3
3
x
ln x
1
1
1
b) f ( x) 

, 1  x  3  L   1  f ' ( x) 2 dx   1  ( x  ) 2 dx   1  ( x 2  
)dx 
1
1
1
2
4
4x
2 16 x 2
1
3
3
3
1
1 2
1 2
1
2
(
x


)
dx

(
x

)
dx

( x  )dx
2
1


1
1
2 16 x
4x
4x
2
x
1
9 1
1
1
 (  ln x) 13  (  ln 3)   4  ln 3
2 4
2 4
2
4
20) ) A)
i)
For dx, the integral needs to be broken up into two pieces as the top equation changes

0
1
ii)
1
( x  1)dx   ( x  1) 2 dx
0
For dy, first solve the equation for x:
do not use
Area=

 y
y  x 1  x  y 1
since the y-values are positive.
and
y  ( x  1) 2  x  1   y .
 x  y 1
1
0
[( y  1)  ( y  1)]dy
B) X
First find the points of intersection:
a) For dx,

x 2  2  x 2  x  1
1
1
[( 2  x 2 )  x 2 ]dx
b) For dy, first solve for x:
y  x2  x   y . y  x2  2  x   2  y
But
Area=

1
0
2
( y  ( y )) dy   ( 2  y  ( 2  y )) dy
1
21) a)
f ( x)  e 4 x  f ( 4) ( x)  256e 4 x .
Thus
error 
The maximum of
256e 4 x
for
0 x4
is
246e16
256e16 (4)5
180(6) 4
c)
f ( x) 
1
2
 f ( 2 ) ( x) 
1 x
(1  x) 3
The maximum of
2
(1  x) 3
2(4) 3
2(4) 3
error 
 0.001  n 
 103.3  104
12(n) 2
12(0.001)
for
0 x4
is
2
2
(1  0) 3