Calculus II
Math 142 Fall 2008
Professor Ben Richert
Exam 2
Solutions
A few integrals:
Z
1
1
sec3 (u) du = tan(u) sec(u) + ln | sec(u) + tan(u)| + C
2
2
Z √
p
2au − u2
a−u
−1
2
du = 2au − u + a cos
+C
u
a
√
√ Z √
√
a + bu − a √
a + bu
du = 2 a + bu + a ln √
√ + C if a > 0
u
a + bu + a √
Z
2u2 − 1
u 1 − u2
u cos−1 u du =
cos−1 u −
+C
4
4
(1)
(2)
(3)
(4)
Z
Problem 1. (10pts) Show how to compute
Z
Solution. We use integration by parts,
f (x) = ln x
f 0 (x) = 1/x
Thus
ln x
dx.
x3/2
f (x)g 0 (x) dx = f (x)g(x) −
Z
f 0 (x)g(x) dx, with the table:
−2
x1/2
1
g 0 (x) = 3/2
x
Z
Z
Z
−2 ln x
−2
−2 ln x
1
−2 ln x
−2
ln x
dx
=
−
dx
=
+
2
dx =
+ 2 1/2 + C.
3/2
1/2
3/2
1/2
3/2
1/2
x
x
x
x
x
x
x
g(x) =
Z
Problem 2. (15pts) Show that
√
10
√
5
3
x
√
dx = 7/6.
2
( x − 1)3
Solution. We use inverse trig substitution. Let x = sec θ whence the lie is that dx = sec θ tan θ dθ. So
Z √10
Z x=√10
Z x=√10
x3
sec3 θ sec θ tan θ
sec4 θ tan θ
√
√
√
dθ
dx =
dθ =
√
√
√
( x2 − 1)3
( sec2 θ − 1)3
( tan2 θ)3
5
x= 5
x= 5
Z x=√10
Z x=√10
sec4 θ tan θ
sec4 θ
=
dθ
=
dθ
√
√
(tan θ)3
tan2 θ
x= 5
x= 5
Z x=√10
Z x=√10
sec2 θ sec2 θ
(tan2 θ + 1) sec2 θ
=
dθ
=
dθ.
√
√
tan2 θ
tan2 θ
x= 5
x= 5
Now let u = tan θ, whence the fib is that du = sec2 θ dθ and we obtain:
Z
√
x= 10
√
x= 5
(tan2 θ + 1) sec2 θ
dθ =
tan2 θ
Z
√
x= 10
√
x= 5
(u2 + 1)
du =
u2
Z
√
x= 10
√
x= 5
1
1 + 2 du
u
x=√10
x=√10
1 1 =u− = tan θ −
.
u √
tan θ √
x= 5
x= 5
p
Using the usual triangle, we find that tan θ = x2 − 1 so we have


√10 q
q√
p
√
1
1
1

= ( 10)2 − 1 − q √
x2 − 1 − √
−  ( 5)2 − 1 − q √
x 2 − 1 √5
2
2
( 10) − 1
( 5) − 1
√
√
1
1
1
1
7
4− √
= 9− √ −
=3− −2+ = .
3
2
6
9
4
Another method is to start by taking u = x2 − 1, which works out nicely.w
Z
Problem 3. (15pts) Demonstrate how to compute
x+1
dx.
(x2 + 1)(x − 1)
Solution. We do expansion by partial fractions. Write
x+1
Ax + B
C
= 2
+
,
(x2 + 1)(x − 1)
x +1
x−1
then clear denominators to obtain the equation
x + 1 = (Ax + B)(x − 1) + C(x2 + 1).
Setting x = 1 yields
1 + 1 = (A + B)(1 − 1) + C(12 + 1) = (A + B)(0) + 2C = 2C,
or C = 1. Thus
x + 1 = (Ax + B)(x − 1) + (x2 + 1),
or
−x2 + x = (Ax + B)(x − 1).
Setting x = 0 yields
0 = (A(0) + B)(0 − 1) = −B,
or B = 0. Thus
−x2 + x = (Ax + 0)(x − 1) = Ax2 − Ax,
so A = −1. Now we have
Z
x+1
dx =
(x2 + 1)(x − 1)
Z −x
1
+
x2 + 1 x − 1
Z
dx = −
x
dx +
x2 + 1
Z
1
dx
x−1
ln |x2 + 1|
+ ln |x − 1| + C
2
where I am using two facts which we know from class:
Z
x
ln |x2 + a|
dx
=
+C
x2 + a
2
and
Z
1
dx = ln |x + a| + C.
x+a
=−
Problem 4. (20pts) Determine whether each integral is convergent or divergent. Show how to evaluate those that are
convergent.
Z 4
4
√ dx.
(a–10pts)
x
0
Solution. This is an improper integral, so:
Z 4
Z 4
√
√
√
√ 4
4
4
√ dx = lim
√ dx = lim 8 x = lim (8 4 − 8 b) = 16 − lim 8 b = 16.
x
x
b
b→0+
b→0+
b→0+ b
b→0+
0
√
We conclude
√ that the
√ integral converges to 16. Note that we used that x is continuous from the right, and thus
that lim 8 b = 8 0 = 0.
b→0+
∞
Z
(b–10pts)
1
2
1 + sin (e
√
x
cos x
)
dx.
Solution. Note that 1 + sin2 (ecos x ) ≥ 1 for all x ≥ 1 since sin2 (ecos x ) = (sin(ecos x ))2 ≥ 0 for all x ≥ 1. Thus
1 + sin2 (ecos x )
1
√
0≤ √ ≤
x
x
Z ∞
1
√ dx diverges by the P -test, implies by the
for all x ≥ 1. This fact, along with the observation that
x
1
Z ∞
1 + sin2 (ecos x )
√
Comparison Test that
dx diverges.
x
1
Problem 5. (15pts) Use the trapezoidal rule to estimate the net distance traveled by a snail from t = 0 to t = 1 if the snail
3
e
has velocity v(t) = et in inches per hour at time t. Your estimate should be have error at most
.
20
Z 1
3
Solution. Note that net change is the integral of rate. So we want to estimate
et dt ≈ Tn .
0
3
3
3
(1 − 0)3 K
where K is such that K ≥ |f 00 (x)| on [0, 1]. Here f 0 (x) = 3t2 et and f 00 (x) = 6tet + 9t4 et .
We know that |ETn | ≤
2
12n
3
3
3
Of course, 6tet + 9t4 et ≥ 0 on [0, 1] since each of the component functions is positive. Also, 6t, 9t4 , and et are increasing
3
on [0, 1] (simply observe that each of their derivatives 6, 32t3 , and 3t2 et are positive on [0, 1]), so that the maximum value
3
3
of f 00 (t) on [0, 1] occurs at t = 1, that is 6tet + 9t4 et ≤ 6e + 9e = 15e on [0, 1]. Thus we may take K = 15e, and conclude
5e
5e
e
5e20
e
15e
=
. In order that |ETn | ≤
we need n such that
≤
, that is, such that n2 ≥
= 25.
that |ETn | ≤
12n2
4n2
20
4n2
20
4e
Here n = 5 does the trick. Now we have
T5 =
with ∆x =
∆x
(v(t0 ) + 2v(t1 ) + 2v(t2 ) + 2v(t3 ) + 2v(t4 ) + v(t5 ))
2
1
, and
5
So
Z
0
inches with error at most
1
3
et dt ≈ T5 =
e
.
20
x0
=
0
x1
=
1/5
x2
=
2/5
x3
=
3/5
x4
=
4/5
x5
=
5/5
2
2
2
2
2
1 03
e + 2e(1/5) + 2e(2/5) + 2e(3/5) + 2e(4/5) + e(5/5)
10