Download Exam 1 Solutions

Exam 1 Solutions
Math 231
February 22, 2007
1.
(a) Write the general formula for integration by parts. (3 pts)
Z
Z
SOLUTION: udv = uv − vdu
Z
(b) Compute sin(2x)e2x dx. (8 pts)
SOLUTION: Use integration by parts twice. u = sin 2x, dv = e2x dx so du = 2 cos 2x and
v = 12 e2x . This gives
Z
Z
1
2x
2x
sin(2x)e dx = sin(2x)e − cos(2x)e2x dx.
2
We now do integration by parts again with u = cos 2x, dv = e2x dx so du = −2 sin 2x. This
gives
Z
Z
1
1
2x
2x
2x
2x
sin(2x)e dx = sin(2x)e −
cos(2x)e − − sin(2x)e dx .
2
2
Solving for the integral in question we get
Z
1 1
1
2x
2x
2x
sin(2x)e dx =
sin(2x)e − cos(2x)e
+ C.
2 2
2
2.
(a) Draw the reference triangle for the substitution 4x = 5 sin θ. (3 pts)
√
SOLUTION:This is a right triangle with sides 4x, 5, and hypotenuse 16x2 + 25.
Z
1
p
(b) Evaluate
dx. (8 pts)
x (ln x)2 − 9
1
SOLUTION: We use the trigonometric solution ln x = 3 sec θ, dx = 3 sec θ tan θ dθ. This
x
makes the new integral
Z
Z
Z
Z
1
3 sec θ tan θ
3 sec θ tan θ
3 sec θ tan θ
p
√
√
dθ =
dx =
dθ =
dθ.
2
2
3 tan θ
9 sec θ − 9
9 tan θ − 9
x (ln x)2 − 9
Using the fact on the first page we complete the integral:
Z
sec θ dθ = ln | sec θ + tan θ| + C
1
Using the reference triangle we finish the problem:
Z
ln x p(ln x)2 − 9 1
p
dx = ln +
+ C.
2
3
3
x (ln x) − 9
Z
3.
(a) Evaluate
sin4 (x) dx. (8 pts)
SOLUTION: We use the half angle formula sin2 x = 21 (1 − cos 2x).
Z
Z 4
sin (x) dx =
2
Z 1 1
1
1
2
(1 − cos 2x)
dx =
− cos 2x + cos 2x dx
2
4 2
4
Now use the half angle formula cos2 2x = 21 (1 + cos 4x) to get
Z
Z
(b) Evaluate
1
1 1
sin (x) dx =
− cos 2x + (1 + cos 4x) dx
4 2
8
x sin 2x x sin 4x
= −
+ +
+C
4
4
8
32
4
Z tan3 x sec x dx. (8 pts)
SOLUTION: We use the Pythagorean identity tan2 x + 1 = sec2 x.
Z
Z
3
tan x sec x dx =
Z
=
Z
=
tan2 x tan x sec x dx
(sec2 x − 1) sec x tan x dx
(sec3 x − sec x) sec x tan x dx
We use the u substitution u = sec x, du = sec x tan x dx and the integral becomes:
Z
4.
1
1
(u3 − u) du = u4 − u2 + C = sec4 x − sec2 x + C.
2
2
(a) Find the general form of the partial fraction decomposition of
find the actual values of A, B, etc.). (3pts)
SOLUTION:
A
B
Cx + D
+
+ 2
2
x + 2 (x + 2)
x +1
2
x2 − 2
(do not
(x2 + 4x + 4)(x2 + 1)
4x2 + 3x + 2
. (6 pts)
x3 + 2x2 + x
SOLUTION: The partial fraction decomposition has the form
(b) Find the partial fraction decomposition of
B
C
A
+
+
.
x x + 1 (x + 1)2
This leads to the system of equations
A+B =4
2A + B + C = 3
A=2
which yields the solution A = 2, B = 2, C = −3, so the partial fraction decomposition is
2
2
3
.
+
−
x x + 1 (x + 1)2
4x2 + 3x + 2
dx. (4 pts)
x3 + 2x2 + x
SOLUTION:
Z
(c) Evaluate
Z
Z
5. Evaluate
√
4x2 + 3x + 2
dx =
x3 + 2x2 + x
Z
2
3
2
+
−
dx
x x + 1 (x + 1)2
3
+C
= 2 ln |x| + 2 ln |x + 1| +
x+1
1
dx. Please work out any trigonometric substitutions that may be neces−x2 − 6x − 5
sary. (10 pts)
SOLUTION: First we complete the square to get −x2 − 6x − 5 = 4 − (x + 3)2 Now use the trig
substitution x + 3 = 2 sin θ dx = 2 cos θ dθ.
Z
Z
Z
1
2 cos θ
√
p
dx =
dθ = 1 dθ = θ + C
−x2 − 6x − 5
4 − 4 sin2 θ
We then use the reference triangle to finish the integral.
Z
1
x+3
√
dx = arcsin
+ C.
2
−x2 − 6x − 5
Z
8
6. Determine if
x−2/3 dx converges or diverges and evaluate the integral if it converges. (10 pts)
−8
SOLUTION: This problem has an infinite integrand at x = 0, so we must separate the integral:
Z 8
Z 0
Z 8
−2/3
−2/3
x
dx =
x
dx +
x−2/3 dx
−8
−8
3
0
We then must use directional limits and we get
Z
8
x
−2/3
Z
0
8
Z
−2/3
dx +
x−2/3 dx
−8
0
8
a
1/3 1/3 = lim− 3x + lim+ 3x dx =
−8
x
a→0
−8
1/3
= lim− 3a
b→0
b
1/3
− 3(−8)
+ lim+ 3(8)1/3 − 3(b)1/3
b→0
a→0
1/3
= −3(−8)
1/3
+ 3(8)
= −3(−2) + 3(2) = 12.
Thus the integral converges.
Z
7. Determine the convergence of
∞
x−2 ln x dx and evaluate the integral if it converges. (9 pts)
1
SOLUTION: This improper integral requires integration by parts as the first step. We select u =
ln x, dv = x−2 dx, and we get du = x−1 dx, v = −x−1 . We now have
Z ∞
Z ∞
ln x ∞
−2
x ln x dx =
−
−x−2 dx
1
x
1
1
ln x a 1 a
= lim
− 1
a→∞ x 1
x 1 1
ln a ln 1
−
−
−
= lim
a→∞ a
1
a 1
=1
Thus the integral converges. The limit used the fact that was given on the front of the page that
ln x
lim
= 0.
x→∞ x
4