Math 106a
Exam 2
11/5/10
Solutions
Z
1. (a)
sec3 x tan3 x dx
Start by splitting off one factor of sec x tan x, use a trigonometric identity to write the tangent
function in terms of secant, then substitute u = sec x.
Z
Z
sec3 x tan3 x dx = sec2 x tan2 x sec x tan x dx
Z
=
Z
=
=
(sec4 x − sec2 x) sec x tan x dx
(u4 − u2 ) du
1 5
5u
(since tan2 x = sec2 x − 1)
(where u = sec x and du = sec x tan x dx)
− 31 u3 + C =
1
5
sec5 x −
1
3
sec3 x + C
Z
(b)
θ sec θ tan θ dθ
To start, apply integration by parts with
u=θ
=⇒
du = dθ
dv = sec θ tan θ dθ
=⇒
v = sec θ
Z
Z
θ sec θ tan θ dθ = θ sec θ −
Z
(c)
sec θ dθ = θ sec θ − ln | sec θ + tan θ| + C
3x + 4
dx
(x − 2)(x2 + 1)
Use partial fractions:
Z
Z 3x + 4
A
Bx + D
dx =
+ 2
dx
(x − 2)(x2 + 1)
x−2
x +1
This gives 3x + 4 = A(x2 + 1) + (Bx + D)(x − 2)
Z
therefore


x = 2 =⇒ A = 2




x = 0 =⇒ D = −1




x = 1 =⇒ B = −2
2
−2x − 1
+ 2
dx
x−2
x +1
Z
Z
Z
2
2x
1
=
dx −
dx
−
dx
x−2
x2 + 1
x2 + 1
3x + 4
dx =
(x − 2)(x2 + 1)
Z (let w = x − 2 and dw = dx; & let u = x2 + 1 and du = 2x dx)
Z
=2
1
dw −
w
Z
1
du −
u
Z
x2
1
dx
+1
= 2 ln |w| − ln |u| − arctan x + C
= 2 ln |x − 2| − ln |x2 + 1| − arctan x + C
1
2. The ellipse to the right is given by x2 + 4y 2 = 1.
Solve for y to find the function that represents the
top half of the ellipse:
p
x2 +4y 2 = 1 ⇐⇒ 4y 2 = 1−x2 ⇐⇒ y = 12 1 − x2 .
An integral which represents the top half of the ellipse is given by Z
1p
1
1 − x2 dx
2
−1
Use trigonometric substitution with x = sin θ for − π2 ≤ θ ≤
π
2,
then dx = cos θ dθ.
Likewise, our substitution gives
θ = arcsin x.
To change limits of integration:
x = −1 ⇒ θ = arcsin(−1) = − π2
x = 1 ⇒ θ = arcsin(1) = π2
Use sin θ = x to construct a
triangle with the opposite side
of length x and hypotenuse of
length 1.
The triangle gives:
p
cos θ = 1 − x2 .
1
2
1
Z
p
Z
p
1 π/2 √ 2
1 − sin2 θ cos θ dθ =
cos θ cos θ dθ
2 −π/2
−π/2
1
2
Z
π/2
1 − x2 dx =
1
2
Z
π/2
=
−1
cos2 θ dθ =
−π/2
1
4
Z
(since cos2 θ = 1 − sin2 θ)
π/2
(1 + cos 2θ) dθ
−π/2
(since cos2 θ = 12 (1 + cos 2θ))
π/2
1
1
π 1
π 1
1
θ + sin 2θ
=
+ sin π − − + sin(−π)
=
4
2
4
2
2
2
2
−π/2
=
Z
3. Evaluate
3
Z
3
6
6
1 π π
+
=
4 2
2
π
4
1
dx.
(4 − x)2
Z 6
1
1
1
dx
+
dx
since
is
undefined
at
x
=
4
2
2
(4 − x)2
3 (4 − x)
4 (4 − x)
Z b
Z 6
1
1
dx + lim+
dx (since each integral is improper)
= lim−
2
2
a→4
b→4
3 (4 − x)
a (4 − x)
1
dx =
(4 − x)2
Z
4
(let u = 4 − x then du = − dx)
4−b
−2
Z 4−b
Z −2
1
1
−2
−2
= lim−
(−u ) du + lim+
(−u ) du = lim−
+ lim+
u
u
a→4
b→4
a→4
b→4
1
4−a
1
4−a
1
1
1
= lim−
− 1 + lim+ − −
4−b
2 4−a
b→4
a→4
1
1
1
But lim
− 1 = ∞ and lim − −
= ∞.
4−b
2 4−a
b→4−
a→4+
Z 6
1
Therefore,
dx diverges.
2
3 (4 − x)
2
4. Consider the function f (x) = e−2x .
(a) The 3rd -order Taylor polynomial, P3 (x), of f (x) centered at x = 0 may be found as follows:
f (x) = e−2x
f 0 (x) = −2e−2x
f 00 (x) = 4e−2x
f 000 (x) = −8e−2x
P3 (x)
=⇒
=⇒
=⇒
=⇒
=
f (0) + f 0 (0)(x − 0) +
=
1 − 2x +
f (0) = e0 = 1
f 0 0) = −2e0 = −2
f 00 (0) = 4e0 = 4
f 00 (0) = −8e0 = −8
f 000 (0)
f 00 (0)
(x − 0)2 +
(x − 0)3
2!
3!
4 2
8
x − x3
2!
3!
4
Therefore, P3 (x) = 1 − 2x + 2x2 − x3 .
3
1
(b) Use P3 (x) to find an estimate for .
e
1
−1
1
Note: f ( 2 ) = e = , so we can estimate this value by calculating P3 ( 21 ), sinceP3 (x) ≈ f (x)
e
2
3
1
1
1
1
4 1
1
for values of x near 0. Therefore, P3
approximates .
= 1−2
+2
−
=
2
2
2
3 2
3
e
(c) We can use Taylor’s Theorem to approximate the error of our estimate from part (b) on the
interval [0, 12 ]. Recall that error bounds for estimates using the 3rd -order Taylor Polynomial may
be determined using:
K4
|f (x) − P3 (x)| ≤
|x − 0|4 .
4!
To find K4 , notice that since f (4) (x) = 16e−2x is decreasing it attains its maximum when x = 0.
So, |f (4) (x)| ≤ |16e−2(0) | = 16 = K4 on [0, 21 ]. Likewise, |x − 0|4 is greatest on [0, 12 ] when x = 21 .
4
1
16 1
Therefore, |f (x) − P3 (x)| ≤
− 0 =
.
4! 2
24
Z ∞
1
5. (a) Recall from class:
dx converges for p > 1 and diverges otherwise. Thus, we know the area
xp
1
Z ∞
1
under 1/x for x > 1 is infinite since
dx diverges. Likewise, the area under 1/x2 for x > 1
x
1
Z ∞
1
is finite since
dx converges. You may think of finding the area between the two functions,
x2
1
for x > 1, as the result of removing a finite area from an infinitely large area, therefore, the result
must be infinite as well. (Or, if you prefer a more rigorous solution,
the shaded area
Z ∞ notice that
1
1
1
1
between y =
and y = 2 for x > 1 is given by the integral
−
dx which diverges
x
x
x x2
1
b
Z b
1
1
1
1
ln b +
since lim
− 2 dx = lim ln x +
= lim
− (0 + 1) = ∞.)
b→∞ 1
b→∞
b→∞
x x
x 1
b
Z ∞
Z ∞
1
1
(b) i. Note:
dx converges and 0 ≤ f (x) ≤ 2 for all x ≥ 1, therefore
f (x) dx converges
x2
x
1
1
by comparison.
Z ∞
1
1
ii. Note:
dx diverges and 0 ≤
≤ g(x) for values of x greater than approximately 2.2
x
x
1
Z ∞
(see graph), therefore
g(x) dx diverges by comparison.
1
Z ∞
1
1
iii. Note: 2 ≤ h(x) ≤ , therefore it is impossible to tell if
h(x) dx converges given the
x
x
1
information provided.
3
6. A 12-inch bar that is clamped at both ends is to be subjected to an increasing amount of stress until
it snaps. Let X be the distance from the left end at which the break occurs. Suppose that X has a
probability density function given by:
x x

1−
if 0 ≤ x ≤ 12
12
f (x) = 24
0
otherwise
(a) The probability that the break point occurs between 2 and 4 inches from the left end of the bar
is given by:
4
Z 4
Z 4 x
1
x2
1 x2
x3
5
x
1−
dx =
x−
dx =
−
=
24
12
24
12
24
2
36
27
2
2
2
(b) The expected break point of the bar is given by:
Z
∞
Z
0
−∞
Z
Z
−∞
=
12
0 dx +
x
0
12 Z
1
24
x2 −
0
∞
xf (x) dx
12
0
0
=
Z
xf (x) dx +
xf (x) dx +
xf (x) dx =
−∞
12
Z
x3
12
x
x2
−
24 288
Z
∞
dx +
0 dx
(refer to definition of f above)
12
dx
12
x4
1 x3
−
= 6
=
24 3
48 0
The expected break point of the bar is 6 inches from the left end.
Z
∞
7. Consider
4
3 + sin θ
dθ.
θ3
We have: − 1 ≤
≤1
for all values of θ
2 ≤ 3 + sin θ ≤ 4
for all values of θ
Thus, 0 ≤
sin θ
3 + sin θ
4
≤ 3
θ3
θ
for θ > 0
Consider the comparison integral,
Z
4
∞
4
dθ = lim
b→∞
θ3
b
Z
4θ
4
−3
b
2
dθ = lim − 2
b→∞
θ 4
b
2
2
1
= lim − 2 + 2 = .
b→∞
b
4 4
8
Z
Therefore,
4
∞
3 + sin θ
dθ converges by comparison. In fact,
θ3
4
Z
4
∞
3 + sin θ
1
dθ ≤ .
θ3
8