Math 132
Fall 2006 Final Exam
2
⌠
2
1. Calculate 
 9x
⌡
0
a) 20
f) 40
b) 24
g) 44
3
1 + x dx.
c) 28
h) 48
d) 32
i) 52
e) 36
j) 56
Solution: i
> with(student):
> J := Int(9*x^2*sqrt(1+x^3),x = 0 .. 2);
2
⌠
2
J := 
 9x
⌡
1 + x 3 dx
0
> K := changevar(1+x^3 = u, J, u);
9
⌠
K :=  3
⌡
u du
1
> antiderivative := int(integrand(K), u);
(3 / 2)
antiderivative := 2 u
> Answer = simplify( subs( u = 9, antiderivative) - subs( u =
1, antiderivative) );
Answer = 52
x
⌠
2. Let F( x ) = 

⌡
t3 − 2 dt. Calculate the derivative D( F )( 3 ) of F at 3.
2
a) 3
b) 4
f) 3 − 3
g) 4 − 5
c) 5
h) 5 − 6
d) 6
e) 7
i) 6 − 7
j) 7 − 2 2
Solution: c
> F := (x) -> Int(sqrt(t^3-2),t = 2 .. x);
⌠
F := x → 

⌡
x
2
t 3 − 2 dt
> D(F)(x);
# By the Fundamental Theorem of Calculus (i.e., no integral
is calculated)
x3 − 2
> D(F)(3);
25
3. Calculate
1
⌠
x+4


 ( x + 1 ) ( x + 2 ) dx .

⌡
0
4
a) ln 
3
8
b) ln 
3
 16 

c) ln
 3 
 25 

d) ln
 3 
 28 

e) ln
 3 
 25 

f) ln
 9 
 32 

g) ln
 9 
 35 

h) ln
 6 
 35 

i) ln
 9 
 36 

j) ln
 5 
Solution: g
> J := Int((x+4)/(x+1)/(x+2),x = 0 .. 1);
1
⌠
x+4

J := 
dx

 (x + 1) (x + 2)
⌡
0
> processedIntegrand := convert( integrand(J), parfrac, x);
processedIntegrand :=
3
−
2
x+1 x+2
> antiderivative := int(processedIntegrand, x);
antiderivative := 3 ln( x + 1 ) − 2 ln( x + 2 )
> answer := simplify(subs(x = 1, antiderivative) - subs(x = 0,
antiderivative));
answer := 5 ln( 2 ) − 2 ln( 3 )
> combine(answer, ln);
 32 
ln 
9
2
⌠
2

 5x +6
4. Calculate 
dx.

2

 (2 + x ) x
⌡
1
a)
π
+ ln( 2 ) b)
4
f) 4 ln( 2 )
π
+ 2 ln( 2 )
4
g) 8 ln( 2 )
π
+ ln( 3 )
4
h) ln( 3 )
c)
d) ln( 2 )
e) 2 ln( 2 )
i) 2 ln( 3 )
j) ln( 6 )
Solution: f
> J := Int( (5*x^2+6)/(2+x^2)/x, x = 1 .. 2);
2
⌠
 5 x2 + 6
J := 
dx

2
 (2 + x ) x
⌡
1
> processedIntegrand := convert( integrand(J), parfrac, x);
processedIntegrand :=
2x
+
3
2 + x2 x
> antiderivative := int(processedIntegrand, x);
antiderivative := ln( 2 + x2 ) + 3 ln( x )
> answer := simplify(subs(x = 2, antiderivative) - subs(x = 1,
antiderivative));
answer := 4 ln( 2 )
1
⌠


5. Calculate 



⌡
0
a) π
b)
f) 2 π
g)
π
2
3π
2
x
2
1−x
2
dx.
π
3
2π
h)
3
c)
d)
i)
π
4
3π
4
e)
j)
π
6
5π
6
Solution: d
> J := Int(x^2/sqrt(1-x^2),x = 0 .. 1);
⌠

J := 


⌡
1
x2
1 − x2
dx
0
> K := student[changevar](x=sin(t), J, t);
#Note: denominator of K is cos(t)
π
2
⌠
 sin( t )2 cos( t )
K := 
dt


1 − sin( t )2
⌡
0
> int(sin(t)^2, t = 0 .. Pi/2);
π
4
6. A radioactive substance has a half-life equal to ln( 8 ) years. If m(t) is the mass of the
substance at time t, measured in years, then what is the logarithmic derivative of m(t) ?
a)
1
2
1
f) −
2
1
3
1
g) −
3
b)
1
c)
d)
2
h) −
1
i) −
2
1
e)
3
1
j) −
3
1
2 2
1
2 2
Solution: g
> k := ln(2)/ln(8); #Solve for decay constant in terms of
half-life
k :=
ln( 2 )
ln( 8 )
> simplify( diff( ln(A*exp(-k*t) ) , t) );
-1
3
> #Note: -ln(2)/ln(8) = -ln(2)/ln(2^3) = -ln(2)/(3*ln(2)) =
-1/3
⌠

7. Calculate 



⌡
1
a) π
b)
3
1
1+x
π
2
2
dx.
c)
π
3
d)
π
4
e)
π
6
f)
5π
6
g)
5π
12
h)
2π
3
i)
3π
4
j)
π
12
Solution: j
> antiderivative := int(1/(1+x^2),x);
antiderivative := arctan( x )
> simplify(subs(x = sqrt(3), antiderivative) - subs(x = 1,
antiderivative));
π
12
x
8. Let f( x ) = ( 1 + 2 x ) . What is D( f )(1) ? (The derivative of f(x) at x = 1) ?
a) 2 ln( 2 ) + 1
f) ln( 3 ) + 2
b) 3 ln( 3 ) + 2
g) ln( 2 ) + 1
c) ln( 2 ) + 2
h) ln( 3 ) + 1
d) ln( 3 ) + 3
i) 3 ln( 2 ) + 1
e) ln( 2 ) + 3
j) 3 ln( 3 ) + 1
Solution: b
> f := x -> (1+2*x)^x;
f := x → ( 1 + 2 x )x
> D(f)(1);
3 ln( 3 ) + 2
9. Consider the following three statements about a series
∞
∑ a with positive terms:
n=1 n
I: The series diverges because
lim a = 1.
n
n→∞
II: The series diverges because
∞
n
lim
= 1 and ∑ b diverges.
b
n=1 n
n→∞ n
a
III: The series diverges because
lim a
n
n→∞
1
 
n
= 1.
For each statement, determine whether the reasoning is correct ( Y ) or incorrect ( N ).
a) I: Y, II: Y,
b) I: Y, II: Y,
III: Y
III: N
c) I: Y, II: N, III: Y
d) I: Y, II: N, III: N
e) I: N, II: Y, III: Y
f) I: N, II: Y, III: N
g) I: N, II: N, III: Y
h) I: N, II: N, III: N
i) Wrong answer
j) Bonus wrong answer
Solution: b
The reasoning of (I) is correct by the Divergence Test.
The reasoning of (II) is correct by the Limit Comparison Test.
The hypothesis of (III) is the inconclusive condition in the Ratio Test. In fact,
∞
the convergence of the series
∑n
n=1
1
2
, a series whose terms satisfy the
stated hypothesis, shows that the reasoning must be wrong
∞
∑
a with positive terms:
10. Consider the following three statements about a series
n=1 n
I: The series converges because lim a = 0.
n
n→∞
∞
II: The series converges because a < b and ∑ b diverges.
n
n
n=1 n
a
III: The series converges because
n
lim
< 1.
a
n→∞ n+1
For each statement, determine whether the reasoning is correct ( Y ) or incorrect ( N ).
a) I: Y,
b) I: Y,
c) I: Y,
d) I: Y,
e) I: N,
f) I: N,
g) I: N,
h) I: N,
II: Y,
II: Y,
II: N,
II: N,
II: Y,
II: Y,
II: N,
II: N,
III: Y
III: N
III: Y
III: N
III: Y
III: N
III: Y
III: N
i) Wrong answer
j) Bonus wrong answer
Solution: h
The reasoning is wrong in all parts. The reasoning of (I) is most likely due to an incorrect
understanding of the Divergence Test. The harmonic series is a counterexample.
The reasoning of (II) results from an incorrect application of the Comparison test.
1
The example of an = and bn = 2 shows that the reasoning is false .
n
The reasoning of (III) results from an incorrect application of the Ratio Test in
∞
which the ratio has been inverted. The divergent series
∑2
n
is a counterexample.
n=1
11. Consider the three series
∞ 5n n
,
I: ∑
n
!
n=1
II:
∞
1
∑
,
n
ln
(
n
)
n=3
and
III:
∞
∑
n=0
n
1
 
3
1+n
 13 


 9 
and the statements
( C ) The series converges
( D ) The series diverges
For each series, decide which of statements, (C) or (D), is correct.
a) I: C, II: C, III: C
b) I: C, II: C, III: D
c) I: C, II: D, III: C
d) I: C, II: D, III: D
e) I: D, II: C, III: C
f) I: D, II: C, III: D
g) I: D, II: D, III: C
h) I: D, II: D, III: D
i) Wrong answer
j) Bonus wrong answer
Solution: c
Applying the Ratio Test to series I leads to a limit of 0, which is less than 1. Hence
convergence.
> a := n -> n*5^n/n!;
n 5n
a := n →
n!
> ratio := simplify( a(n+1)/a(n) );
5
ratio :=
n
> limit(ratio, n = infinity);
0
Applying the Integral Test to series II leads to the conclusion of divergence.
> Int(1/x/ln(x),x=3..infinity) = int(1/x/ln(x),x=3..infinity);
∞
⌠
1


dx = ∞

 x ln( x )
⌡
3
∞
Applying the Limit Comparison Test with
∞
∑b =∑
1
n
n=1
n=1
n
 10 


 9 
leads to the conclusion of
convergence.
> a := n -> n^(1/3)/(1+n^(13/9));
b := n -> 1/n^(10/9);
a := n →
n
(1 / 3)
1+n
( 13 / 9 )
1
b := n →
( 10 / 9 )
n
> Limit(a(n)/b(n), n = infinity) = limit(a(n)/b(n), n =
infinity);
lim
n→∞
n
( 13 / 9 )
1+n
( 13 / 9 )
12. Consider the two series
∞ ( −1 )n
∞ ( − 1 ) n n3
I: ∑
and II: ∑
ln
(
n
)
n
n=3
n=0
2
and the statements
i: The series converges absolutely
=1
ii: The series converges conditionally
iii: The series diverges
For each series, decide which of statements ($), (&), (@) is correct.
a) I: i, II: i
b) I: i, II: ii
c) I: i,
II: iii
d) I: ii, II: i
e) I: ii, II: ii
f) I: ii,
II: iii
g) I: iii, II: i
h) I: iii, II: ii
i) I: iii, II: iii
j) Wrong answer
Solution: d
13. Consider the two series
n
∞
∞
10
I: ∑
and II: ∑
n=0
n = 0 10n + n2
and the statements
( C ) The test establishes convergence
( D ) The test establishes divergence
( X ) The test is not conclusive.
n
3
3
n
Apply the Ratio Test to series I and the Root Test to series II. For each, decide
which of statements, (C), (D), (X), is correct.
a) I: C, II: C
b) I: C, II: D
c) I: C, II: X
d) I: D, II: C
e) I: D, II: D
f) I: D, II: X
g) I: X,
h) I: X , II: D
i) I: X , II: X
j) Wrong answer
II: C
Solution: g
∞ ( n + 3 ) xn
14. Let f( x ) = ∑
. What is f ''' ( 0 ) ?
2
n = 1 n (n + 1)
a) 1
f) 1/3
b) 2
g) 1/2
c) 3
h) 2/3
d) 6
i) 4/3
e) 12
j) 1/6
Solution: a
> a := n -> (n+3)/(n^2*(n+1));
n+3
a := n →
n (n + 1)
2
> simplify(3!*a(3));
1
> p := x -> sum((n+3)*x^n/(n^2*(n+1)),n = 1 .. 10);
10
p := x →
( n + 3 ) xn
∑n
n=1
2
(n + 1)
> (D@@3)(p)(0);
1
15. Let T( x ) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 be the degree 5 Taylor polynomial
of
f ( x ) = x2 e
a) 1/5
f)
- 1/5
( −2 x )
centered about 0. What is T( 1 ) ?
b) 2/5
g)
- 2/5
c) 1/3
d) 2/3
h) - 1/3
i)
e) 4/15
- 2/3
j) - 4/15
Solution: h
> S := series(x^2*exp(-2*x), x = 0, 6);
S := x2 − 2 x3 + 2 x4 −
4
x5 + O( x6 )
3
> p := unapply(convert(S,polynom),x);
p := x → x2 − 2 x3 + 2 x4 −
4
3
x5
> p(1);
-1
3
∞
16. Calculate the interval of convergence of
∑
n=0
a) ( - 5 , 1 ]
f) [ - 5 , 1 )
b) [ - 5 , 1 ]
g) ( - 5 , 1 )
c) ( - 1 , 5 ]
h) [ - 1 , 5 )
( − 1 )n n ( x + 2 )n
(n + 1) 3
2
d) [ - 1 , 5 ]
i) ( - 1 , 5 )
n
.
e) [ - 2 , 2 ]
j) ( −∞ , ∞ )
Solution: a
Because x + 2 = x − ( −2 ), the center is c = −2.
The next calculation determines the radius of convergence.
> a := n -> n/(n^2+1)/(3^n);
a := n →
n
( n2 + 1 ) 3n
> R := limit(a(n)/a(n+1), n = infinity);
R := 3
The left endpoint is x = c − R, or x = −2 − 3, or x = −5. When this value
∞
is substituted in
∑
n=0
( −1 )n n ( x + 2 )n
( n2 + 1 ) 3n
∞
, the result is
∑n
n=0
n
2
+1
.
This series is divergent by applying the Limit Comparison Test with the harmonic series.
Thus, -5 is out.
The right endpoint is x = c + R, or x = −2 + 3, or x = 1. When this value
∞
is substituted in
∑
n=0
( −1 )n n ( x + 2 )n
( n2 + 1 ) 3n
∞
, the result is
∑
n=0
( −1 )n n
n2 + 1
.
This series is convergent by the Alternating Series Test.
Thus, 1 is in.
3
17. What is the coefficient of ( x − 4 ) in the Taylor series of
1
32
1
f) −
32
a)
3
64
3
g) −
64
b)
1
128
1
h) −
128
c)
3
256
3
i) −
256
d)
x with base point 4?
1
512
1
j) −
512
e)
Solution: e
> simplify(subs(x=4,diff(sqrt(x),x$3))/3!);
1
512
> series(sqrt(x), x=4, 5); #alternative
2+
1
4
(x − 4) −
1
64
( x − 4 )2 +
1
512
5
( x − 4 )3 −
16384
( x − 4 )4 + O( ( x − 4 )5 )
0.1
⌠
2
18. The Maclaurin series of sin( x ) is used to approximate 
 42 sin( x ) dx with an
⌡
0
error
( −6 )
. The calculation uses only as many terms as are deemed necessary for the
less than 10
required accuracy by the Alternating Series Test. What is the approximation?
2
a) 0.012
f) 0.017
b) 0.013
g) 0.018
c) 0.014
h) 0.019
d) 0.015
i) 0.020
e) 0.016
j) 0.021
Solution: c
> series(42*sin(x^2),x=0,16);
42 x2 − 7 x6 +
7
20
> p := convert( %, polynom );
x10 −
p := 42 x2 − 7 x6 +
1
120
x14 + O( x16 )
7 x10
20
−
x14
120
> int(subs(x=t, p), t = 0 .. x);
14 x − x +
3
7
7 x11
220
−
x15
1800
> subs(x=0.1, 14*x^3);
0.014
> evalf(Int(42*sin(x^2),x = 0 .. 0.1));
0.01399990000
19. What is the coefficient of x
a) −
5
27
b) −
14
81
c) −
3
4
27
in the Maclaurin series of
d) −
10
81
e) −
1
9
x
?
3+2x
f)
5
27
g)
14
81
h)
4
27
10
81
i)
1
9
j)
Solution: h
> series(x/(3+2*x),x=0,8);
1
x−
2
4
x2 +
8
x3 −
3
9
27
81
> convert( %, polynom);
x
3
−
2 x2
9
+
4 x3
27
x4 +
−
16
243
8 x4
x5 −
+
81
32
729
16 x5
243
−
x6 +
64
2187
32 x6
729
+
x7 + O( x8 )
64 x7
2187
> coeff(%,x^3);
4
27
20. What is the coefficient of x
a) −
f)
5
27
5
27
b) −
g)
14
81
14
81
c) −
h)
3
in the Maclaurin series of
4
27
d) −
10
81
e) −
i)
10
81
j)
4
27
1
1
 
3
(1 + x)
?
1
9
1
9
Solution: b
> binomial(-1/3,3); #From Newton's Binomial Series
-14
81
> series(1/(1+x)^(1/3), x = 0, 5); #direct calculation
1−
1
x+
2
x2 −
14
x3 +
35
x4 + O( x5 )
3
9
81
243
> product(-1/3-k,k=0..2)/3!; #calculation of binomial
coefficient
-14
81