Calculus I
Exam 3, Fall 2002, Answers
1. Integrate:
5 x 1 dx Answer. Let u x 4x 5 !du 4 x 1 dx. Then
1
u x 4x 5 x 4x 5 x 1 dx u du C
4
16
16
b)
tan x 1 sec xdx Answer. Let u tan x du sec xdx. Then
u tan x tan x 1 sec xdx u 1 du u C tanx 3
3
a)
x
4
3
4x
3
4
3
4
2
3
3
4
3
4
4
C
2
2
2
2
dy
dx
2. Solve the differential equation:
3
2
xy2
y2
3
C
0 Answer. We can separate the variables, getting
y
which integrates to
2
dy
xdx
x2 C 1
y
x 2 C 1
0
2 C
y
so that
2
1
2
Substituting the initial conditions:
which has no solution. Thus there is no function of this type satisfying this initial condition. Recall, however,
that when we separated variables, we divided by y 2 ; this can only be done if y 0. Thus the alternative y 0
remains, and provides the solution: the function which is identically zero.
3. Calculate the definite integrals:
1 dx
x 3 4
3 4
Answer. x x 4 4 0
3 2
3 2
3
b)
cos x sin x dx
Answer. Let u sin x du cos xdx. Then when x 0 u cos x sin x dx 4
a)
x2
3x
0
3
2
4
0
3
2
π 2
0
0 and when x
π 2
0
1
udu
0
1
2
π 2 u 1, and we have
4. Find the area of the region in the first quadrant bounded by the curves y
x1
x and y
4
4x 2 .
Answer. The second curve is always above the first, and both leave the first quadrant when x 1. Thus the
area is
1
1
x2
5
1
Area
4 4x2
x x2 dx
4 x 3x2 dx
4x
x3 0
2
2
0
0
5. The region in the first quadrant bounded by the curves y
is the volume of the solid so produced?
x 2 and x
1 is rotated about the y-axis. What
Answer. We sweep out the volume along the x-axis, as x ranges from 0 to 1. The volume of the shell of width
dx at the point x is dV 2π xydx 2π x 3 dx. Thus the volume is
Volume
1
0
2π x3 dx
π
2