Calculus I
Final Exam, Spring 2003, Answers
1. Find the integrals:
a)
esinx 2 cos xdx
Answer. Let u
cos xdx. Then
e
e
cos xdx sin x du
sinx 2
Alternatively, let v
b)
x x
u 2
esinx dv
du
e2u du
e2u
2
e2 sinx
2
C
esinx cos xdx, so that
esin x 2 cos xdx
esinx esinx cos xdx
vdv
v2
2
C
C
esin x
2
2
C
1dx
x 1 so that x u 1 and du dx. Then
2 2 x x 1dx u 1 u du u
u du 5 x 1 3 x 1 C Answer. Let u
1 2
2. Integrate
t2
3 2
1 2
5 2
3 2
2 dt
t2
1 t
1 t 2 B t 1 t 2 C t 1 t 1 t 2
Answer. . We expand the function in partial fractions. The roots are -1,1,2, so we write
2 t 1 t 1 t 2 Equate the numerators at the roots.
1 A 2 3 so A t 1 :
12
t 1 : 1 B 2 1 so B 4
t 2 : 2 C 4 1 so C 3
t2
t2
1 t
A
B
C
At
2
2
2
1
6
2
2
This gives us
3. Integrate
t2
dt 2
16 ln t 1 t2
1 t
t 1
1
ln t 1 2
1
6
dt
1
2
dt
4
3
dt
1 t
4
ln t 2 C 3
t
2
x ln xdx
Answer. We integrate by parts so as to get rid of the ln term: u ln x dv xdx, so that du
and
x2
x2 dx x2
x2
x ln xdx
ln x
ln x
C
2
2 x
2
4
dx x v
x 2 2,
4. A certain compound transforms from state A to state B at a (per minute) rate proportional to the concentration of B in the mixture:
dcA
02cB
dt
where cA and cB are the concentrations of A and B respectively (and, assuming no other material is present,
cA cB 1). If at time t 0 the mixture is 90% in state A how long will it take to be 10% A?
Answer. Substitute cB
1
cA in the differential equation, and separate variables, obtaining
dcA
1 cA
02dt ln 1 c 02t C exponentiating to 1 c Ke Solve for K using c 9 when t 0, getting K 1. Now solve for c :
c 1 1e (Of course this makes sense so long as c 0; once A is gone, the process stops.) Now set c 1 and solve
for t: 1e 9, so
ln 9
t
02 109 86 minutes Integrate both sides and exponentiate:
02t
A
A
A
A
02t
A
A
02t
A
5. Find the limit. Show your work.
Answer. At x ln x
1 sin π x
a) lim
x
1, both numerator and denominator are zero, so l’Hôpital’s rule applies:
x
b) lim
lim 1 x 1 π cos π x π
ln x
1 sin π x
lim
lH
x 1
1
xex
x 0 e2x
Answer. Again both numerator and denominator are zero at x
lim
x
3x6 7x4
∞ 2 x3
12
c) lim
0, so
lim e xe 1 2e
xex
x 0 e2x
x
lH
x
2x
x 0
1
2
3
2
since the factors have the same degree.
6. Do the integrals converge? If so, evaluate:
∞
a) Answer.
0
xe x dx
lim
A ∞ 0
A
xe x dx
lim
xe e lim e A 1 1 1 x
A ∞
x
A
0
A
A ∞
∞
b)
2
dx
x ln x
25
Answer. Let u
ln x du
dx x. Then
dt x ln x which converges to 1 24 ln2
as A A
24
du
25
ln 2 u
25
2
u24 A
1
24
A
ln 2
1
ln 2
ln A1 24
24
∞.
24
7. Do the series converge or diverge? Give a valid reason for your answer.
n 1
n 1
Answer. The series diverges by comparison with the p-test with p ∞
a)
∑
2
n 1
3
1: the denominator is only of degree 1
more than the numerator.
∞
b)
ln n
n
1 2
∑
n
Answer. The series converges by comparison with a geometric series:
ln n
2n
2n 2
2n
12 n
2 1.
n! 1 c) ∑
n 1 !
series converges by comparison with the p-test with p 2. Divide both numerator and denomAnswer. The
inator by n! :
n! 1 1 2
!
"
n 1!
n 1
and 1
∞
2
2
n 1
2
1 2
n!
n 1! 2
n!
2
2
2
since the numerator is bounded by 2.
8. Find the vertices of the conic given by the equation 4x 2
1 y
Answer. Complete the square:
4
y
2
8x
4y
12
4x 1 y 2 4 4 0 or
This gives the standard form
x 31 y 122 1 This is a hyperbola with center at (-1,-2) and axis the line x 1. Setting x the vertices:
y 2
12 1 or y 2 # 12 Thus the vertices are at 1 2 2 3 and 1 2 2 3 .
4x
2
2x
2
4y
12
2
0
2
2
12
2
1 gives the y coordinates of
2
9. Find the area of the region enclosed by the curve given in polar coordinates by r
the segment of the x axis between x 2 and x 2e 2π .
θ
2e θ 0
2π and
PSfrag replacements
200
50
$ 200$ 0
50
$ 100
$ 150
$ 200
$ 250
$ 300
$ 350
0
200
400
600
800
1000 1200
Answer. From the diagram we see that the area is
Area
1
2
2π
0
r2 d θ
2π
e2θ d θ
2
0
e2θ
2π
0
e4π
10. a) Find the general solution of the homogeneous differential equation y
Answer. The roots of the equation r 2
3r
2
yh
% %& 3y% 2y
0.
0 are 1,2. Thus the general solution is
Aex
Be2x
b) Find a particular solution of the homogeneous differential equation y
1
% %' 3y% 2y
sin x.
Answer. We use the method of undetermined coefficients. Try a solution of the form y
A cos x
Bsin x:
3 A sin x Bcos x 2 A cos x Bsin x sin x leading to the equations A 3B 2A 0 ( B 3A 2B 1. The solutions are B 1 10 A 3 10, so
a particular solution is
y 0 3 cos x 0 1 sin x A cos x
Bsin x
p