Solution: i
Math 132
Spring 2013 Final Exam
x
?
ÿ 2 ln( 1 + t2 + t3 )
ÿ
1. Let F( x ) = ÿ
dt. Calculate F' (4), the derivative of F(x) at x = 4.
ÿ
ln
(
1
+
)
t
ÿ
?
0
a) 0
f) 5
b) 1
g) 6
c) 2
h) 7
d) 3
i) 8
e) 4
j) 9
Solution: i
> F := (x) -> Int(2*ln(1+t^2+t^3)/ln(1+sqrt(t)), t = 0 .. x);
x
?
ÿ 2 ln( 1 + t2 + t3 )
F := x ¡ú ÿ
dt
ÿ
ÿ ln( 1 + t )
?
0
> D(F)(x);
#
# This is calculated using the Fundamental Theorem of Calculus
- without any integration
2 ln( 1 + x2 + x3 )
ln( 1 +
x)
> derivative := D(F)(4);
Answer = simplify( derivative );
2 ln( 81 )
derivative :=
ln( 1 + 4 )
Answer = 8
e
?
ÿ 12 1 + 8 ln( x )
2. Calculate ÿ
dx.
ÿ
x
ÿ
?
1
a) 21
b) 22
c) 23
d) 24
e) 25
f) 26
g) 27
h) 28
i) 29
j) 30
Solution: f
> J := Int(12*sqrt(1+8*ln(x))/x,x = 1 .. exp(1));
#
# Given integral
e
?
ÿ 12 1 + 8 ln( x )
J := ÿ
dx
ÿ
x
ÿ
?
1
> K := changevar(u = 1 + 8*ln(x), J, u);
#
# Integral with same value obtained by substitution u =
1+8*ln(x), du = (1/x)*dx
9
?
ÿ 3 u
K := ÿ
du
ÿ
ÿ 2
?
1
> K = value(K);
9
?
ÿ 3 u
ÿ
du = 26
ÿ
ÿ 2
?
1
Verification using Maple's built-in integrator
> J = int(12*sqrt(1+8*ln(x))/x,x = 1 .. exp(1));
e
?
ÿ 12
ÿ
ÿ
ÿ
?
1
2
?
ÿ 4x+3
3. Calculate ÿ
dx .
ÿ
ÿ x (x + 1)
?
1
1 + 8 ln( x )
x
dx = 26
a) ln( 2 )
b) ln( 3 )
c) ln( 4 )
d) ln( 6 )
e) ln( 8 )
f) ln( 9 )
g) ln( 12 )
h) ln( 16 )
i) ln( 18 )
j) ln( 24 )
Solution: g
> with(student):
> J := Int((4*x+3)/x/(x+1),x = 1 .. 2);
2
?
ÿ 4x+3
J := ÿ
dx
ÿ
ÿ x (x + 1)
?
1
> processedIntegrand := convert( integrand(J), parfrac, x);
#
# Partial fraction form of integrand
processedIntegrand :=
3
+
1
x x+1
> antiderivative := int(processedIntegrand, x);
#
# Indefinite integral of given definite integral
antiderivative := 3 ln( x ) + ln( x + 1 )
> answer := simplify(subs(x = 2, antiderivative) - subs(x = 1,
antiderivative));
#
# Value of given integral
answer := 2 ln( 2 ) + ln( 3 )
> Displayed_answer = combine(answer, ln);
#
# Combine logarithms to get listed answer: Uses p*ln(u) =
ln(u^p) and ln(x)+ln(y) = ln(x*y)
ln( 12 )
Verification using Maple's built-in integrator:
> Int((4*x+3)/x/(x+1),x = 1 .. 2) = int((4*x+3)/x/(x+1),x = 1 ..
2);
2
?
ÿ 4x+3
ÿ
dx = 2 ln( 2 ) + ln( 3 )
ÿ
ÿ x (x + 1)
?
1
> testeq( 2*ln(2)+ln(3) = ln(12) );
true
4. Calculate
a) ¦Ð +
1
4
1
?
ÿ
ÿ 4 x2 + x + 4
ÿ
dx.
ÿ
2
ÿ
ÿ
ÿ ( x2 + 1 )
?
0
b) ¦Ð +
1
2
c) ¦Ð + 1
¦Ð
f) 2 ¦Ð + 1
g)
4
d) 2 ¦Ð +
¦Ð
+1
h)
4
1
e) 2 ¦Ð +
4
¦Ð
+2
i)
2
1
2
¦Ð
j)
+1
2
+2
Solution: a
> J := Int( (4*x^2+x+4)/(x^2+1)^2, x = 0 .. 1);
1
?
ÿ 4 x2 + x + 4
ÿ
J := ÿ
dx
2
ÿ
ÿ ( x2 + 1 )
?
0
> processedIntegrand := convert( student[integrand](J), parfrac,
x);
#
# Partial fraction form of given integrand
x
processedIntegrand :=
2
+
4
2
x +1
( x2 + 1 )
> antiderivative := int(processedIntegrand, x);
#
# Antiderivative for integrand of given integral
antiderivative := ?
1
+ 4 arctan( x )
2 ( x2 + 1 )
> Answer := simplify(subs(x = 1, antiderivative) - subs(x = 0,
antiderivative));
1
Answer := + ¦Ð
4
Verification using Maple's built-in integrator:
> J = value(J);
1
?
ÿ 4 x2 + x + 4
1
ÿ
d
x
=
+¦Ð
ÿ
2
ÿ
4
2
ÿ (x + 1)
?
0
¦Ð
2
?
(2 x)
¦Ð
5. Given that ÿ
cos( x ) dx = e ? 2, what is
ÿ 5e
?
0
a) e¦Ð ? 3
b) e¦Ð ? 1
f) 2 e¦Ð ? 3
g) 2 e¦Ð ? 1
¦Ð
e +1
¦Ð
h) 2 e + 1
c)
¦Ð
2
?
ÿ 5 e( 2 x ) sin( x ) dx
ÿ
?
0
d) e¦Ð + 2
e) e¦Ð + 3
i) 2 e¦Ð + 2
j) 2 e¦Ð + 3
?
Solution: h
> J := Int(5*exp(2*x)*sin(x),x=0..Pi/2);
#
# J is the definite integral to be evaluated
¦Ð
2
?
(2 x)
J := ÿ
sin( x ) dx
ÿ 5e
?
0
> K := intparts(J,exp(2*x));
#
# Applying integration by parts to J with u = exp(2*x) and dv =
5*sin(x)*dx
# The expression K has the same value as J
................
................
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