Calculus II
Math 142 Fall 2013
Professor Ben Richert
Exam 2
Solutions
Z
Problem 1. (10pts) Compute
tan5 x sec3 x dx. You must show your work, of course, but for this problem
you are not required to give an explanation in English.
Solution. We use the usual trig substitution techniques. First we save tan x sec x to be the Lie and clear the
decks of the remaining tan x, thus:
Z
Z
Z
5
3
2
2
2
tan x sec x dx = (tan x) sec x tan x sec x dx = (sec2 −1)2 sec2 x tan x sec x dx.
Now let u = sec x whence the Lie is that du = tan x sec x dx. So
Z
Z
2
2
2
(sec −1) sec x tan x sec x dx = (u2 − 1)2 u2 du
Z
=
(u6 − 2u4 + u2 ) du =
u5
u3
sec7 x
sec5 x sec3 x
u7
−2 +
+C =
−2
+
+ C.
7
5
3
7
5
3
Z
Problem 2. (10pts) Compute
x3 ln x dx You must show your work, of course, but for this problem you
are not required to give an explanation in English.
Solution. We use integration by parts. So
Z
Z 3
Z
x4
x4
x4
x4
1 x4
x
ln x −
dx =
ln x −
dx =
ln x −
+C
x3 ln x dx =
4
x 4
4
4
4
16
Z
Z
f g 0 dx = f g − f 0 g dx
f = ln x g = x4 /4
f 0 = 1/x g 0 = x3
x−1
dx. You must show your work, of course, but for this problem
x2 − 4x − 5
you are not required to give an explanation in English.
Z
Problem 3. (10pts) Compute
Solution. The integrand is a rational function and there is no need to long divide, so we factor the denominator and then use the method of partial fractions:
x−1
x−1
A
B
=
=
+
.
x2 − 4x − 5
(x + 1)(x − 5)
x+1 x−5
Clearing denominators yields
x − 1 = A(x − 5) + B(x + 1) = (A + B)x + (−5A + B)
and equating coefficients gives
1
=
A+B
−1
=
−5A + B.
Subtracting the second equation from the first gives 2 = 6A, so A = 1/3 and hence B = 1 − A = 2/3. Thus
Z Z
1/3
2/3
1
2
x−1
dx
=
+
dx = ln |x + 1| + ln |x − 5| + C.
x2 − 4x − 5
x+1 x−5
3
3
Z
1
Note the use of the liver bit
dx = ln |x − a| + C.
x−a
Z
∞
x+1
dx. You must show your work, of course, but for this problem
2 + 2x
x
1
you are not required to give an explanation in English.
Problem 4. (10pts) Compute
du
= (x + 1) dx. Also x = 1 =⇒
2
Solution. We use u-substitution. Let u = x2 + 2x whence the Lie is that
u = 3 and x = t =⇒ u = t2 + 2t. Thus
Z
∞
1
x+1
dx = lim
t→∞
x2 + 2x
Z
t
1
x+1
1
dx = lim
t→∞ 2
x2 + 2x
Z
t2 +2t
3
1
du
u
t2 +2t
1
1
ln |u| = lim
ln(t2 + 2t) − ln 3 = ∞
t→∞ 2
t→∞ 2
3
= lim
2
2
(we
Z ∞ know the last bit since t + 2t → ∞ and hence ln(t + 2t) → ∞ as t → ∞). We conclude that
x+1
dx diverges.
x2 + 2x
1
Problem 5. (10pts) Compute
Z p
−x2 + 2x dx. You must show your work, of course, but for this problem
you are not required to give an explanation in English.
Solution. First we complete the square
p
−x2 + 2x =
p
−(x2 − 2x) =
q
p
− (x − 1)2 − 1 = 1 − (x − 1)2 .
So
Z p
−x2
+ 2x dx =
Z p
1 − (x − 1)2 dx.
Now let u = x − 1 whence the Lie is that du = dx and
Z p
Z p
1 − (x − 1)2 dx =
1 − u2 du.
Now we use an inverse trig substitution. Let u = sin t whence the Fib (having already Lied) is du = cos tdt.
Thus
Z p
Z √
Z p
1 − u2 du =
1 − sin2 t cos t dt =
cos2 t cos t dt
Z
=
cos2 t dt =
Z
1
1
(1 + cos 2t) dt =
2
2
1
1
1
t + sin 2t + C =
t + 2 sin t cos t + C.
2
2
2
Note the use of the half angle formula and the trig identity sin 2t = 2 sin t cos t.
Now to divest ourselves of t, consider that t lives in a triangle with
u
1
←→
u = sin t
t
p
1 − u2
and so
p
1
1 −1
sin (u) + u 1 − u2 + C
t + 2 sin t cos t + C =
2
2
p
1
=
sin−1 (x − 1) + (x − 1) 1 − (x − 1)2 + C.
2
Z
√
Problem 6. (10pts) Compute
1 + ex dx. You must show your work, of course, but for this problem
1
2
you are not required to give an explanation in English.
√
Solution. We can get out of trouble by mentioning what a wonderful function ex is. Let u = 1 + ex whence
x
√
e
dx. Of course, u = 1 + ex implies that u2 = 1 + ex or ex = u2 − 1, we
the Lie is that du = √
x
2 1+e
rewrite the Lie as
√
2 1 + ex du
= dx
ex
or
2u du
= dx.
u2 − 1
So
Z
Z
√
2u2
x
1 + e dx =
du.
2
u −1
Long dividing, we find
2u2
2
2
=2+ 2
=2+
2
u −1
u −1
(u + 1)(u − 1)
2
Now we use the method of partial to break up
. So
(u + 1)(u − 1)
A
B
2
=
+
,
(u + 1)(u − 1)
u+1 u−1
and clearning denominators gives
2 = A(u − 1) + B(u + 1) = (A + B)u + (−A + B)
whence equating coefficients yields
0
=
A+B
2
= −A + B.
Adding these two equations gives 2 = 2B, so B = 1 and (from the first equation) A = −B = −1. Thus
Z
Z Z 2u
2
1
1
du
=
2
+
du
=
2
−
+
du
u2 − 1
(u + 1)(u − 1)
u+1 u−1
√
√
√
2u − ln |u + 1| + ln |u − 1| + C = 2 1 + ex − ln( 1 + ex + 1) + ln( 1 + ex − 1) + C.
Z
1
Note the use of the liver bit
dx = ln |x − a| + C.
x−a
Z
Problem 7. (15pts) Consider the integral
1
p
1 + x4 dx.
0
Z
(a–5pts) Use Simpson’s rule with n = 4 to write down a sum which approximates
1
p
1 + x4 dx. Do not
0
simplify, of course. (For this part of the problem, you need not give an explanation in English.)
Solution. We take ∆x =
1−0
1
= whence
4
4
x0 =
0
x1
=
1/4
x2
=
2/4
x3
=
3/4
x4
=
4/4.
So
∆x
(f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 ))
3
p
p
p
p
1 p
=
1 + (0)4 + 4 1 + (1/4)4 + 2 1 + (2/4)4 + 4 1 + (3/4)4 + 1 + (4/4)4 .
12
S4 =
(b–10pts) Suppose that your roommate, having studied for 30 hours straight, flips out. Instead of sleeping,
Z 1p
1 + x4 dx with n = 100. He also
he computes (by hand!) the Midpoint Rule estimate for
0
p
2x2 (3 + x4 )
computes that the second derivative of 1 + x4 is
. Give an upper bound for the error
(1 + x4 )3/2
in your roommate’s estimate. Need I say it? Don’t simplify.
Z b
Solution. We know that using Mn to estimate
f (x) dx results in error EM (n) and furthermore
a
that if K is some number such that |f 00 (x)| ≤ K on [a, b], then
K(b − a)3
.
24n2
p
2x2 (3 + x4 )
In this situation the second derivative of 1 + x4 is
. Note that on [0, 1]
(1 + x4 )3/2
|EM (n)| ≤
d 2
x = 2x ≥ 0,
dx
d
(3 + x4 ) = 4x3 ≥ 0,
dx
and
d
3
(1 + x4 )3/2 = (1 + x4 )1/2 4x3 ≥ 0,
dx
2
that is, each of x2 , 3 + x4 , and (1 + x4 )3/2 are increasing on [0, 1]. So we can maximize x2 and
1
(3 + x4 ) by evaluating at x = 1 and maximize
by evaluating at x = 0 (because doing
(1 + x4 )3/2
so minimizes (1 + x4 )3/2 ). Thus,
x2 ≤ 1,
3 + x4 ≤ 4,
and
1
≤1
(1 + x4 )3/2
on [0, 1], and it follows that
0≤
1
2x2 (3 + x4 )
= 2x2 (3 + x4 )
≤ 2(1)(4)(1) = 8
(1 + x4 )3/2
(1 + x4 )3/2
on [0, 1]. We conclude that
2
2x (3 + x4 ) (1 + x4 )3/2 ≤ 8,
on [0, 1], that is, we can take K = 8. So
|EM (100)| ≤
whatever that is.
8(1 − 0)3
,
24(100)2