test i solution

7/28/2019 Test i Solution

1/4

Sultan Qaboos University

Department of Mathematics and Statistics

Math2108 Fall 2011

Test I Time: 60 minutes

Name:. . . . . . . . . . KEY . . . . . . . . . Section: . . . . . Number. . . . . . . .

Important Instructions

Write your name, ID # and Section # on the front cover of your answer booklet. In all questions, you must show your complete, mathematically correct and neatly written solution. You are NOT allowed to share calculators or any other material during the test. Cellular phones are NOT allowed to be used for any purpose during the test. You should NOT ask the invigilator any questions about the exam.

Q1: (5 points)

Sketch the region bounded by the curves of y = |x| and y = 2 x2, then find its area.Solution: First, we find the intersection between the curves y = |x| and y =2 x2 by solving

x = 2 x2 and x = 2 x2.

From the fist equation, we obtain x = 1,2. x = 1 is the valid solution. Fromthe second equation, we obtain x = 1, 2. x = 1 is the valid solution.

Now, the area is given by

A =

1

1

(2 x2 |x|)dx =

0

1

(2 x2 + x)dx +

1

0

(2 x2 x)dx = 76

+7

6=

7

3.

Remark: Because |x| and 2 x2 are even functions, we can say

A =11

(2 x2 |x|)dx = 21

0(2 x2 x)dx = 2

7

6 =7

3 .

However, we cannot conclude that based on the sketch only.

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Q2: Sketch the region enclosed by the curves of y =

x, y = 6 x and y = 0, then usecylindrical shells to find the volume of the solid generated by revolving the region aboutthe x-axis. (5 points)

Solution: First, we find the intersection between the curves by solving

x = 6 x or x = 36 12x + x2.We obtain (x 4)(x 9) = 0 and the valid solution is x = 4.

Now, the method of cylindrical shells gives us

V = 2

2

0

y(6 y y2) dy = 2

2

0

(6y y2 y3) dy = 323

.

Q3: The region enclosed by the triangle with vertices (0, 0), (0, h), (r, 0), r > 0, h > 0 is rotatedabout the y-axis. Use integration to find the volume of the generated solid. (5 points)

Solution: First, we need to find the equation of the line passing through the

points (0, h), (r, 0). The equation is given by

y =h

r

(r

x).

Next, we find the volume as follows:

V = 2

r

0

xh

r(r x) dx = 2 h

r

r

0

rx x2 dx = 13

r2h.

Q4: The integral (3 points)

1

0

2x3

1 +

9

16x4 dx

represents the surface area of a solid. Answer each of the following:

(i) Find the function that produces the solid.

Solution: The function that produces the solid is given by f(x) = 14

x3.

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(ii) What is the axis of revolution.

Solution: The axis of revolution is the x-axis.

Q5: Let n 1 and a be fixed constants. Show that (3 points)

xneax dx = 1a

xneax na

xn1eax dx.

Solution: Using integration by parts, we take

u = xn, dv = eaxdx, du = nxn1dx, v =1

aeax.

So, xneax dx =

1

axneax

n

axn1eax dx =

1

axneax n

a

xn1eax dx.

Q6: Evaluate each of the following integrals: (4+5 points)

(i)

e2

1

y2 ln(y) dy (ii)

4

x3 + 4xdx.

Solution: (i) Using integration by parts, we obtain

e2

1

y2 ln(y) dy = ln(y)y3

3

e2

1

e2

1

1

y

y3

3dy,

which implies e2

1

y2 ln(y) dy =2

3e6 1

3

e

1

y2 dy =2

3e6 1

9e6 +

1

9.

Next, we solve (ii). We use partial fraction decomposition to write

4

x3 + 4x=

4

x(x2 + 4)=

A

x+

Bx + C

x2 + 4,

then we find the constants A,B and C. We have

4 = A(x2

+ 4) + (Bx + C)x.

Combine the alike terms to obtain

A = 1, A + B = 0, C = 0.

Thus, A = 1, B = 1, C = 0. Now,

4

x3 + 4xdx =

1

xdx

x

x2 + 4dx

= ln |x| 12

2x

x2 + 4dx

= ln |x| 12

ln(x2 + 4) + c.

3


4/4

Q7: Evaluate each of the following integrals: (5+5 points)

(i)

0

sin5(x)cos4(x) dx (ii)

9 x2 dx.

Solution: (i)

0

sin5(x)cos4(x) dx =

0

sin(x)sin4(x)cos4(x) dx =

0

sin(x)(1 cos2(x))2 cos4(x) dx.

Now, we use substitution. Let cos(x) = y. We obtain

sin(x)dx = dy

and

0

sin5(x)cos4(x) dx = 1

1

(1 y2)2y4 dy =

1

1

(y4 2y6 + y8) dy.

Since the integrand is an even function, we obtain

1

1

(y4 2y6 + y8) dy = 2

1

0

(y4 2y6 + y8) dy = 16315

.

Next, we solve (ii). We use trig substitution. Let x = 3 sin(y). We obtain

dx = 3 cos(y)dy

and 9 x2 dx = 9

cos(y)

1 sin2(y) dy = 9

cos(y)| cos(y)| dy.

We dont know the limits of integration; however, if we restrict the angle to

make | cos(y)| = cos(x),then, we obtain

9

cos2(y) dy =

9

2

(1 + cos(2y)) dy =

9

2(y +

1

2sin(2y)) + c.

Finally, we get x back to obtain

9 x2 dx = 9

2(y +

1

2sin(2y)) + c =

9

2sin1(

x

3) +

1

2x

9 x2 + c.

Total: 40 points

Good Luck

test i solution

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