Math 1B – Calculus II – Fall ’11 – Chapter 7 Problems
Name_______________________
Write all responses on separate paper. Show your work for credit. Don’t use a calculator.
1. Find the volume generated when the region bounded by y = sinx and y = 0 between x = 0 and x = π is revolved
around the y-axis. Recall that an element of volume is given by the shell dV = 2πrhdx.
2. Use integration by parts to find a reduction formula for
3. Evaluate the integral
∫
π /6
0
∫
1
0
x 2 n e3 x dx .
cos ( 3 x ) cos ( 6 x ) dx by
a. Using the identity cos 2θ = 2 cos 2 θ − 1
b. Using the identity cos A cos B = ½ [cos(A – B) + cos (A + B)]
4. A particle moves along a straight line with velocity v(t) = cos(ωt)sin2(ωt).
Find its position function s = f (t) if f (0) = 1.
∫x
5. Find
2
1
dx in terms of a if
−a
a. a < 0
b. a > 0
6. Use trigonometric substitution to prove that
∫
x
0
1 − t 2 dt =
1
x
arcsin ( x ) +
1 − x2
2
2
then interpret both terms on the right side of
the equation in terms of areas in the figure
at right.
4
2
7. Find the average value of f ( x ) = x 4 − x on the interval [0,2].
8. Show the proper way of evaluating the improper integral
∫
∞
1
1
dx
x +1
2
9. Determine whether the improper integral converges or diverges. Use comparison, if necessary.
a.
∫
∞
b.
∫
∞
1
2
1
x2 + 1
1
x3 + 1
dx
dx
Math 1B – Calculus II – Fall ’11 – Chapter 7 Problem Solutions
1. Find the volume generated when the region bounded by y = sinx and y = 0 between x = 0 and x = π is revolved
around the y-axis. Recall that an element of volume is given by the shell dV = 2πrhdx.
SOLN: Using
2
u=x
dv = sin xdx
du = dx
v = − cos x
2
sin
cos |
2
cos
2. Use integration by parts to find a reduction formula for
2
∫
1
0
0
0
0
2
x 2 n e3 x dx .
u = x2n
dv = e3 x dx
SOLN: Using
1
du = 2nx 2 n −1dx v = e3 x
3
∫
3. Evaluate the integral
π /6
0
cos ( 3 x ) cos ( 6 x ) dx by
a. Using the identity cos 2θ = 2 cos 2 θ − 1
∫
π /6
0
cos ( 3x ) ( 2cos 2 ( 3x ) − 1) dx = ∫
π /6
0
π /6
0
=∫
π /6
0
= 2∫
SOLN:
2 cos3 ( 3 x ) dx − ∫
π /6
0
cos ( 3x ) dx
cos ( 3x ) (1 − sin 2 3x ) dx − ∫
π /6
0
cos ( 3x ) dx − 2∫
π /6
0
cos ( 3x ) dx
cos ( 3x ) sin 2 3xdx
(Let u = sin ( 3x ) so that du = 3cos ( 3x ) dx)
π /6
1
sin 3x
2u 3
1 2 1
=
−
= − =
3 0
9 0 3 9 9
b. Using the identity cos A cos B = ½ [cos(A – B) + cos (A + B)]
SOLN:
∫
π /6
0
π /6
π /6
sin ( 3x )
sin ( 9 x )
1 π /6
1 1 1
cos ( 3 x ) cos ( 6 x ) dx = ∫ cos ( −3 x ) + cos ( 9 x ) dx =
+
= − =
2 0
6
18 0
6 18 9
0
4. A particle moves along a straight line with velocity v(t) = cos(ωt)sin2(ωt).
Find its position function s = f (t) if f (0) = 1.
1
SOLN:
1
cos
cos
sin
sin
. Substituting v = sin(ωu) so that dv = ωcos(ωu)du, we have
1
1
1
3
1
sin
3
5. Find
∫x
2
1
dx in terms of a if
−a
a. a < 0
If a < 0 then substitute
tan
so that
arctan
and
√
√
√
tan
√
sec
√
and
√
√
b. a > 0
If a > 0 then we can factor the denominator to get
√
√
√
√
√
2
ln
√
ln
√
√ sec
Note that this can also be done with the substitution,
√
√ integral becomes
√ sec tan
so that
and the
√ √ √
ln csc
which is another, equivalent, expression.
√
cot
ln
√
√
√
6. Use trigonometric substitution to prove that
arcsin
√1
√1
then interpret both
terms on the right side of the equation in terms of areas in the figure.
SOLN: The area in the figure is
Let
sin so that
cos
√1
sin
2
√1
and we have
sin
√1
cos
cos
cos
2
1
arcsin
2
In the diagram,
arcsin
with central angle
is
2
1
, so the area of the unit circle sector
arcsin
triangular region is ½ base*height =
and the area of the
√1
4
2
7. Find the average value of f ( x ) = x 4 − x on the interval [0,2].
SOLN: Average value is
√4
.
Let x = 2sin(θ) so that the integral
,
is
/
16
/
32
/
4
sin
2 cos
sin
sin cos
sin 2
4 sin
√4
1
/
4
2
4
sin
cos
/
32
cos 2
/
32
sin 2
4
sin 2 cos 2
cos 4
8. Show the proper way of evaluating the improper integral
lim
SOLN:
lim
arctan
→
lim
→
arctan |
→
π
arctan 1
9. Determine whether the improper integral converges or diverges. Use comparison, if necessary.
a.
√
√
lim
→
lim
→
lim
→
√
Substitute
/
1
√
ln 1
√2
divergent…but that’s hard. Why not observe that if x > 1 then
√2
so that
b.
√
√
2
1
3
2
lim
b→∞
2
so that
∞ so the integral is
√
√
√
∞?
1
SOLN: Note that
limb→∞
∞1
sec
ln
tan ,
√
1
2
1
√
lim 2
b→∞
so that
2
√
2, so the integral is convergent.
0