Calculus I
Exam 3, Summer 2002, Answers
1. Integrate:
5 x 1 dx Answer. Let u x 3x 5 du 3 x 1 dx. Then
1
x 3x 5 x 1 dx 3
b)
sin x 1 cos xdx Answer. Let u sin x du cos xdx. Then
sin x 1 cos xdx u 1 du a)
x
3
3x
3
2
3
2
3
3
2
u3 du
1 3
x
12
3x
5 C 4
2
2
2
1 3
u
3
u C 1 3
sin x
3
C
sinx
y 1
2 dy 1 x dx. Integrating:
1 x x C y
2
Putting in the initial values we get 1 2 1 1 2 C, so C 2. Putting in this value of C, and solving
for y:
x
y 2 x
2. Solve the differential equation:
dy
1 x y2
dx
Answer. As an equation of differentials, this becomes y
2
2
2
1
2
3. Calculate the definite integrals:
4
a)
x
3
0
3x
1 dx
Answer.
π 2
b)
0
4
x
1 dx x 4
4
3
3x
0
3x2
2
x 4
0
64
24
4
92
sin x cos x dx
Answer.
π 2
sin x cos x dx
0
sin2 x
2
π 2
0
1
2
4. Find the area of the region in the third quadrant bounded by the curves y
x 3 and y
2x
Answer. Draw the graph. The region in question is that in the third quadrant running from x
Since the upper curve is y x 3 , the area is
0
x
2
3
2x
x dx 2
x4
4
x
2
x3
3
164 4 83 0
2
8
3
x2 .
2 to x
0.
PSfrag replacements
30
20
10
3
1
2
0
0
1
2
3
10
20
30
5. The region in the first quadrant bounded by the curves y
is the volume of the solid so produced?
x and y
x is rotated about the y-axis. What
Answer. Draw the graph. Sweeping along the x axis and using the shell method we obtain
1
PSfrag replacements
V
2π x
x x dx 2π
1
x x dx 3 2
2
2π
2
1 5 3
0
30
Sweeping along the y-axis2and using the washer method we first rewrite the curves as x
1
1
0 1
1 1
2π
π y2 π y4 dy π
y2 y4 dy π
V
3 5
15
0
0
2
3
30
12
20
10 1
10
200 8
30
06
04
02
10 0
0
02
04
06
08
1
12
2π
15
y and x
y 2 . Then