Solution: i - Department of Mathematics and Statistics

Math 132 Spring 2013 Final Exam

1.

Let

F( x ) =

x 2 ln( 1 + t2 + t3 )

ln( 1 + t )

dt.

Calculate

F' (4), the derivative of

F(x)

at

x = 4.

0

a) 0 b) 1 c) 2 d) 3 e) 4 f) 5 g) 6 h) 7 i) 8 j) 9

Solution: i

> F := (x) -> Int(2*ln(1+t^2+t^3)/ln(1+sqrt(t)), t = 0 .. x);

F

:=

x

x 2

ln( 1 + t2 ln( 1 +

+ t3 ) t)

dt

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

2. Calculate e 12 1

1 + 8 ln( x ) dx.

x

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

J := e 12

1 + 8 ln( x ) dx

x

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

> K = value(K);

K := 9 3 2 u du

1

9 3 2 u du = 26

1

Verification using Maple's built-in integrator

> J = int(12*sqrt(1+8*ln(x))/x,x = 1 .. exp(1));

e 12

1

1 + 8 ln( x ) dx = 26

x

3.

Calculate

2

4 x(

x x

+ +

3 1)

dx

.

1

a) ln( 2 ) f) ln( 9 )

b) ln( 3 ) c) ln( 4 ) d) ln( 6 ) e) ln( 8 ) 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);

J

:=

2

4 x(

x x

+ +

3 1)

dx

1

> processedIntegrand := convert( integrand(J), parfrac, x);

#

# Partial fraction form of integrand

31 processedIntegrand := +

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

4 x(

x x

+ +

3 1)

dx

=

2

ln( 2 )

+

ln( 3 )

1

> testeq( 2*ln(2)+ln(3) = ln(12) );

true

4. Calculate

1

4 x2 ( x2

+ +

x 1

+ 4 2 )

dx.

0

1 a) +

4 f) 2 + 1

1 b) +

2

g) + 1

4

c) + 1

h) + 2

4

1 d) 2 +

4

i) + 1

2

1 e) 2 +

2

j) + 2

2

Solution: a

> J := Int( (4*x^2+x+4)/(x^2+1)^2, x = 0 .. 1);

J

:=

1

4 x2 + x + 4

(

x2

+

1

2

)

dx

0

> processedIntegrand := convert( student[integrand](J), parfrac,

x);

#

# Partial fraction form of given integrand

x

4

processedIntegrand :=

+

(

x2

+

1

2

)

x2 + 1

> antiderivative := int(processedIntegrand, x);

#

# Antiderivative for integrand of given integral

1

antiderivative := -

+ 4 arctan( x )

2 ( x2 + 1 )

> Answer := simplify(subs(x = 1, antiderivative) - subs(x = 0,

antiderivative));

1 Answer := +

4

Verification using Maple's built-in integrator:

> J = value(J);

1

4 x2 ( x2

+ +

x+4

2

1)

dx

=

1 4

+

0

5.

Given that

2

5

(2 e

x

)

cos( x )

dx

=

e

-

2,

what is

2

5

( e

2

x

)

sin(

x

)

dx

?

0

0

a) e - 3

b) e - 1

c) e + 1

d) e + 2

e) e + 3

f) 2 e - 3

g) 2 e - 1 h) 2 e + 1

i) 2 e + 2 j) 2 e + 3

Solution: h

> J := Int(5*exp(2*x)*sin(x),x=0..Pi/2); # # J is the definite integral to be evaluated

J

:=

2

5

(2

e

x

)

sin(

x

)

dx

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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