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Math 152 – Calculus II Second Midterm Solution November 1, 2006
- Evaluate the following indefinite integrals. Clearly indicate which methods you apply.
(a) (3 points) ∫
(cos x)
5 (sin x)
3 dx
Answer
I =
(cos x)
5 (sin x)
3 dx =
(cos x)
5 (1 − cos
2 x) sin xdx
We substitute u = cos x, so that du = − sin xdx to obtain
I =
−u
5 (1 − u
2 )du = −u
6 /6 + u
8 /8 + C
Substituting back u = cos x we obtain
I = −
(cos x)
6
(cos x)
8
(b) (3 points) ∫ e
√ x
√ x
dx
Answer
We substitute u =
x, so that u
2 = x and 2udu = dx and obtain
e
u
u
2 udu =
2 e
u du = 2e
u
√ x
- Evaluate the following indefinite integrals. Clearly indicate which methods you apply.
(a) (3 points) ∫ x + 1
x^2 − 2 x + 2
dx
Answer
If we complete the square on the denominator we get x
2 − 2 x + 2 = (x − 1)
2
substitute u = x − 1, so that x = u + 1 and dx = du. We get
u + 2
u^2 + 1
du =
u
u^2 + 1
du + 2
u^2 + 1
du =
ln |u
2
Rewriting in terms of x gives:
ln |x
2 − 2 x + 2| + 2 arctan(x − 1) + C
(b) (3 points) ∫ x
cos^2 (x)
dx
Answer
Since
d dx
tan x =
1 cos^2 (x)
we can write:
x
cos^2 (x)
dx =
xd(tan x) = x(tan x) −
tan(x)dx
We have
∫
tan(x)dx =
sin(x)
cos(x)
dx =
cos x
d(cos(x)) = − ln | cos x| + C,
so the whole answer is:
∫ x
cos^2 (x)
dx = x(tan x) + ln | cos x| + C
- (2 points) We approximate the value of an integral
∫ (^) b
a
f (x)dx using the midpoint rule Mn and
using Simpson’s rule Sn. Suppose we have computed that for n = 10 we have ∣ ∣ ∣ ∣
∫ (^) b
a
f (x)dx − M 10
− 1 and
∫ (^) b
a
f (x)dx − S 10
− 1 .
(a) (1 point) How big should we take n to make sure that
∫ (^) b
a
f (x)dx − Mn
∣ ≤^10
− 5 ? Explain.
Answer
The error bound for the midpoint rule is of the form
∫ (^) b
a
f (x)dx − Mn
≤ CM /n
2
where CM is some constant depending on the behaviour of f
′′ (x) for a ≤ x ≤ b but not on
n. It is given we can take CM / 10
2 = 10
− 1 , so CM = 10.
In order to have CM /n
2 ≤ 10
− 5 we need n
2 ≥ 10
6 , so choosing n = 1000 should do the
trick.
(b) (1 point) How big should we take n to make sure that
∫ (^) b
a
f (x)dx − Sn
∣ ≤^10
− 5 ? Explain.
Answer
The error bound for Simpson’s rule is of the form
∫ (^) b
a
f (x)dx − Mn
≤ CS /n
4
where CS is some constant depending on the behaviour of f
′′′′ (x) for a ≤ x ≤ b but not on
n. It is given we can take CS / 10
4 = 10
− 1 , so CS = 1000.
Solving CM /n
4 ≤ 10
− 5 shows we should have n
4 ≥ 10
8 , so n = 100 should do.
- (3 points) Evaluate the following definite integral. Clearly indicate your methods.
− 1
ln(x
2 )dx
Answer
Since ln(x
2 ) has an asymptote at x = 0, this is an improper integral, which we will compute by
− 1
ln(x
2 )dx =
− 1
ln(x
2 )dx +
0
ln(x
2 )dx.
We have (^) ∫ 1
0
ln(x
2 )dx = lim h→0+
h
ln(x
2 )dx.
By partial integration we have
h
ln(x
2 )dx = [x ln(x
2 )]
1 h −
h
xd(ln x
2 ) = (1· 0 −h ln(h
2 ))−
h
x
x^2
· 2 xdx = − 2 h ln(h)−2(1−h).
In order to compute
lim h→0+
h ln(h),
we substitute h = 1/y and let y → ∞. We then get
lim y→∞
− ln(y)
y
which via de l’Hˆopital, converges to 0. Therefore,
lim h→0+
− 2 h ln(h) − 2(1 − h) = − 2
and we see that (^) ∫ 1
0
ln(x
2 )dx = − 2.
Since ln(x
2 ) is an even function, we have
∫ (^) −h
− 1
ln(x
2 )dx =
h
ln(x
2 )dx
so the other improper integral converges to −2 as well. Adding together yields
− 1
ln(x
2 )dx = − 4.
