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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