Calculus II
Math 142 Winter 2011
Professor Ben Richert
Exam 2
Solutions
Problem 1. (10pts) Compute exactly one of the following two integrals. Be sure to indicate clearly which you want graded.
A single sentence explanation is sufficient (for instance, what technique are you using? why did you choose that technique?).
Z
x3
√
(a–10pts)
dx
x2 + 1
Solution. This looks like inverse trig substitution.
So let x = tan θ and lie by saying dx = sec2 θ dθ. You get:
Z
Z
x3
tan3 θ sec2 θ
tan3 θ sec2 θ
√
√
√
dx =
dθ
=
dθ
x2 + 1
tan2 θ + 1
sec2 θ
Z
Z
tan3 θ sec2 θ
dθ = tan3 θ sec θ dθ.
=
sec θ
Now this appears to be a standard trig integral, so let u = sec θ whence the fib is that du = tan θ sec θ dθ, and
Z
Z
tan3 θ sec θ dθ = (sec2 θ − 1)tanθ sec θ dθ
√
3
p
x2 + 1
u3
sec3 θ
−u+C =
− sec θ + C =
− x2 + 1 + C
3
3
3
p
2
Note that we have replaced sec
pθ with x + 1 since x = tan θ, so θ lives in a triangle with opposite side x,
adjacent side 1, and hypotenuse x2 + 1.
Z
1
√
(b–10pts)
dx
x2 1 − x2
Z
=
(u2 − 1) du =
Solution. This (again) looks like inverse trig substitution.
So let x = sin θ and lie by saying dx = cos θ dθ. You get:
Z
Z
cos θ
1
√
p
dx =
dθ
2
2
2
x 1−x
sin θ 1 − sin2 θ
Z
Z
cos θ
1
√
=
dθ =
dθ
2
2
sin2 θ
sin θ cos θ
√
Z
1 − x2
= csc2 θ dθ = − cot θ + C = −
+ C.
x
√
1 − x2
Note that we have replaced cot θ with
since x = sin θ, so θ lives in a triangle with opposite side x,
x
p
hypotenuse side 1, and adjacent side 1 − x2 .
Z 2
ln(x)
√ dx. A single sentence explanation is sufficient (for instance, what technique are you
Problem 2. (10pts) Compute
x
1
using? why did you choose that technique?).
Solution. This appears to be integration by parts, so we proceed with the formula
Z
Z
f g 0 dx = f g + f 0 g dx.
1
We take f = ln(x) and g 0 = √ yielding the table:
x
√
f (x) = ln(x) g(x) = 2 x
1
1
f 0 (x) =
g(x) = √
x
x
and thus
Z
1
2
2 Z 2 2
√ 2 √
√
√
√
√
ln(x)
√ dx = 2 x ln(x) −
√ dx = 2 2 ln(2) − 2 1 ln(1) − 4 x = 2 2 ln(2) − 4 2 + 4.
x
x
1
1
1
Z
2
2x + 2x + 2
dx. A single sentence explanation is sufficient (for instance, what technique
(x + 1)(x2 + 1)
are you using? why did you choose that technique?).
Problem 3. (10pts) Compute
Solution. This begs for the method of partial fractions. The degree of the denominator is larger than that of the numerator,
so we do not do long division. Also x2 + 1 is irreducible (because 02 − 4(1) < 0), so we don’t need to do any factoring.
Now set
2x2 + 2x + 2
A
Bx + C
=
+ 2
(x + 1)(x2 + 1)
x+1
x +1
or clearing denominators:
2x2 + 2x + 2 = A(x2 + 1) + (Bx + C)(x + 1)
=
Ax2 + A + Bx2 + Bx + Cx + C
=
(A + B)x2 + (B + C)x + (A + C)
yielding the system:
A+B
=
2
B+C
=
2
A+C
=
2
Subtracting the first two equations yields:
A−C =0
and adding the result with the third equation gives
2A = 2,
or A = 1, whence it follows that C = 1 and again that B = 1. So
Z Z
1
x+1
2x2 + 2x + 2
dx
=
+
dx
(x + 1)(x2 + 1)
x + 1 x2 + 1
Z
Z
Z
1
x
1
=
dx +
dx +
dx
x+1
x2 + 1
x2 + 1
1
= ln |x + 1| + ln |x2 + 1| + arctan(x) + C
2
where we have used several of the “bits” as described in class.
Problem 4. (10pts) Compute the following two limits:
ln(x)
(a–5pts) lim
x→∞
x
∞
Solution. Here we have ln(x) → ∞ and x → ∞ as x → ∞, so the form is “ ” and all the derivatives behave, thus
∞
we apply L’Hospital’s rule:
1
ln(x)
1
= lim x = lim
= 0.
x→∞
x→∞ 1
x→∞ x
x
lim
(b–5pts) lim x1/x (it’s ok to use your answer from part (a) if it comes up)
x→∞
Solution. We have
1/x
lim x1/x = lim eln(x
x→∞
x→∞
)
= lim e
x→∞
ln(x)
x
= elimx→∞
ln(x)
x
= e0 = 1,
where we use the fact that the limit can pass the e (since ex is continuous), the log rules, and our answer from part
(a).
Z 2
Problem 5. (10pts) Consider the integral
− ln(x) dx. We could compute this directly, but let’s practice approximating.
1
Z 2
(a–3pts) Use Simpson’s rule with n = 4 to estimate
− ln(x) dx. Don’t simplify, of course. This is the only portion of the
1
test for which no English is required: after all, what is there to say?
2−1
= 1/4 and xi = 1 + i∆x, so
4
∆x
S4 =
(− ln(x0 ) − 4 ln(x1 ) − 2 ln(x2 ) − 4 ln(x3 ) − ln(x4 ))
3
1
= − (ln(1) + 4 ln(5/4) + 2 ln(6/4) + 4 ln(7/4) + ln(2))
12
Solution. We have ∆x =
1
(b–7pts) How big must n be to ensure that the error in Sn is at most
?
180(6)3
Solution. We know that if K is such that K ≥ |f 0000 (x)| on [1, 2] then |ESn | ≤
have
K(2 − 1)5
. Taking derivatives, we
180n4
= − ln x
1
f 0 (x) = −
x
1
00
f (x) =
x2
2
f 000 (x) = − 3
x
6
0000
f (x) =
x4
f (x)
6
24
is a decreasing function on [1, 2] (its derivative, after all is − 5 , which is negative of [1, 2]), so it
x4
x
6
6
1
is largest when x = 1, that is K = 6 ≥ | 4 | works. So we solve for n such that |ESn | ≤
≤
, and find
x
180n4
180 · 63
3
6 · 180 · 6
n4 ≥
= 64 . Obviously n = 6 works.
180
Of course,
Problem 6. (10pts) Do exactly one of the following two problems. Indicate clearly which you want graded.
Z ∞
1
(a–10pts) Compute
dx or show that it diverges.
x
ln(x)
2
dx
Solution. We do u-substitution, letting u = ln(x) whence the lie is du =
, so
x
Z ∞
Z a
1
1
dx = lim
dx
a→∞ 2 x ln(x)
x ln(x)
2
Z x=a
x=a
a
1
= lim
du = lim ln |u|
= lim ln | ln(x)|
a→∞
a→∞ x=2 u
a→∞
x=2
2
= lim (ln(ln(a)) − ln(ln(2))) = ∞,
a→∞
so the interval diverges.
Z
(b–10pts) Use the comparison test to determine whether
1
∞
2
sin x
dx converges or diverges.
x3
1
sin2 x
≤ 3 on [1, ∞). Recall that
Solution. Note that 0 ≤ sin2 x ≤ 1 for all real x, so 0 ≤
3
x
x
Z ∞
sin2 x
dx converges by the Comparison Test.
by the P-test. We conclude that
x3
1
Z
1
∞
1
dx converges
x3