test1-fall12 solution

7/28/2019 Test1-Fall12 Solution

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Sultan Qaboos University - College of ScienceDepartment of Mathematics and Statistics

MATH2108 (Calculus II)- Fall 2012 - TEST 1SOLUTIONS

1. (7 points) Sketch and find the area of the region bounded by y = x2 7 andy = x 1.

Solution. The intersection points of the parabola y = x27 and the line y = x1are deduced from the equation x2 7 = x 1, i.e. x2 x 6 = 0. Solving thisquadratic equation, we obtain the roots x = 2 and x = 3. Hence the requiredarea is equal to

A =

32

(x 1) (x2 7)dx

=

32

x x2 + 6dx

=x2

2 x

3

3+ 6x

|32=

9

2 27

3+ 18

4

2+

8

3 12

= 5

2 35

3+ 30

=15 70 + 180

6=

125

6.

2. (7 points) Compute the volume of the solid obtained by revolving the regionbounded by y = x3 1, the y-axis and y = 1 about y = 1.

Solution. The variable to take for calculating the volume is y. The corresponding

radius is y + 1; and the corresponding height is essentially the abscissa value ofthe intersection of the horizontal line with value y and y = x3 1, i.e. height is


2/5

3

y + 1. Therefore the required volume is equal to

V =

+11

2(y + 1)(y + 1)1

3 dy

= 2

+11

(y + 1)4

3 dy

[y + 1 = t, 0 t 2, dy = dt]

= 2

20

t4

3 dt

= 2 37

t7

3 |20

=

6

7 2

7

3

=6

7 22 2 13

=24

721

3 .

If a student uses washers method, then outer radius is 2, inner radius is x3 1 (1) = x3. Hence

V =

32

0

[22

(x3

)2

]dx

=

32

0

[4 x6]dx

= (4x x7

7) | 3

2

0

=24 3

2

7.

3. (6 points) Compute the arc length for the following curve exactly: y = 23

(x2 + 1)3/2

on [1, 4].

Solution. The formula for arc length of a curve gives us that the required arclength is equal to

l =

41

1 + (y(x))2dx.


3/5

Since y(x) = (23

(x2 + 1)3/2) = 23 32 (x2 + 1) 12 2x = 2x(x2 + 1) 12 , we have that

l =

41

1 +

2x(x2 + 1)

1

2

2dx

=41

1 + 4x2(x2 + 1)dx

=

41

1 + 4x4 + 4x2dx

=

41

(1 + 2x2)2dx

=

41

(1 + 2x2)dx

= (x +2

3x3) |41

= 4 +2

3 64 1 2

3

= 3 +2

3 63

= 3 + 2 21= 45.

4. (5+5 points) Evaluate the following integrals:

(a)

120

x tan1(2x)dx

(b)

dxx2x24 , x > 2

Solution. (a). We do integration by parts:

u = tan1(2x) dv = xdx

du =2

1 + 4x2dx v =

x2

2.


4/5

Therefore

120

x tan1(2x)dx = uv |

12

0

120

vdu

=x2

2tan1(2x) |

12

0

120

x2

1 + 4x2dx

=2

288tan1(

6) 0 1

4

120

4x2

1 + 4x2dx

=2

288 tan1(

6) 1

4

120

4x2 + 1 11 + 4x2

dx

=2

288 tan1(

6) 1

4

120

dx +1

4

120

1

1 + 4x2dx

=2

288 tan1(

6) 1

4x |

12

0 +1

8tan1(2x) |

12

0

=2

288 tan1(

6)

48+

1

8 tan1(

6)

=2 63 + 36

288

3; OR

0.026504

(b). We will use the following trigonometric substitution: x = 2 sec . We also willmake use ofx24 = 4 sec2 4 = 4(sec2 1) = 4 tan2 , and dx = 2 tan sec d.Hence

dx

x2

x2 4 =

2tan sec

4sec2 2tan d

=1

4

1

sec d

=

1

4

cos d

=1

4sin + C.

Now, cos = 2x

and so sin =

1 cos2 =

1 ( 2x

)2. Thus

dx

x2

x2 4 =1

4

1 ( 2

x)2 + C.


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5. (5+5 points)

Evaluate the following integrals:

(a)

3

0 tan3 x sec

3

2 xdx.

(b)

x+4x3+3x210xdx

Solution. (a). The substitution to make is u = sec x. Then du = tan x sec xdx,tan2 x = sec2 x 1 and 1 u 2. Thus

30

tan3 x sec3

2 xdx =

30

tan2 x sec1

2 x tan x sec xdx

=

2

1

(u2

1)u

1

2 du

=

21

(u5

2 u 12 )du

=2

7y

7

2 23

y3

2

|21=

2

7 2 72 2

3 2 32 2

7+

2

3

=16

2

7 4

2

3 2

7+

2

3

.

(b). First decompose the given rational function:

x + 4

x3 + 3x2 10x =x + 4

x(x2 + 3x 10) =x + 4

x(x 2)(x + 5) =A

x+

B

x 2 +C

x + 5.

From this one derives that

x + 4 = A(x 2)(x + 5) + Bx(x + 5) + Cx(x 2).

Letting x = 0, one obtains 4 =

10A, thus A =

2

5

.

Letting x = 5, one obtains 1 = 35C, thus C = 135

.

Letting x = 2, one obtains 6 = 14B, thus B = 37

.

Thereforex + 4

x3 + 3x2 10x dx = 2

5

dx

x+

3

7

dx

x 2 1

35

dx

x + 5

= 25

ln |x| + 37

ln |x 2| 135

ln |x + 5| + C.

test1-fall12 solution

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