Selected Solutions, Section 5.2
1. This is good practice in taking left endpoints.
In this case, f (x) = 3 x/2, and the interval is [2, 14]. The Riemann sum using 6
rectangles will use:
Width of each rectangle: (14 2)/6 = 12/6 = 2.
The height of the rec
Exam 1 solutions (M126C)
1. Determine if the sequence converges or diverges. If it converges, nd its limit: an =
(ln(n)2
n
SOLUTION: Use lHospitals rule (twice):
(ln(n)2
2 ln(n) (1/n)
ln(n)
1/n
= lim
= lim 2
= lim 2
=0
n
n
n
n
n
1
n
1
lim
Therefore, the s
Exam 2 Review Solutions
1. State the Fundamental Theorem of Calculus: Let f be continuous on [a, b].
x
If g(x) =
f (t) dt, then g (x) = f (x).
a
b
f (x) dx = F (b) F (a), where F is any antiderivative of f .
a
n
b
2. Give the denition of the denite integ
Selected Solutions, Section 5,1
2. The purpose of this exercise is to be sure we have a little experience with estimating
area using rectangles. The midpoint rule is to evaluate the heights of the rectangles
at the midpoint of each rectangle.
Recall that
Final Exam Review
Calculus II
Sheet 3
1. Determine if the series converges (absolute or conditional) or diverges:
(a)
(1)n ln(n)
n
n=1
SOLUTION: We might rst check to see if it converges absolutely. It will not, by
the integral test. To check that the fun
Review Sheet 1 Solutions
n
i2 =
1. Prove by induction:
i=1
n(n + 1)(2n + 1)
6
SOLUTION:
Prove the rst case: If n = 1, then does 12 =
123
?
6
Yes.
k
k(k + 1)(2k + 1)
.
6
i=1
Use the assumption to prove that the statement is true if n = k + 1. In that case
Final Exam Review
Calculus II
Sheet 2
1. True or False, and give a short reason:
2n+3
(a) The Ratio Test will not give a conclusive result for 3n4 +2n3 +3n+5
TRUE. The ratio test fails for plike series (the limit will be 1). To show convergence, use a dir
Solutions to the Review Questions, Exam 3
1. A cross section of a tank of water is the bottom half of a circle of radius 10 ft, and is 50
ft long. Find the work done in pumping the water over the rim of the tank if it lled to
a depth of 7 feet (set up the
Selected Solutions, Section 5.4
10. Hint: Multiply it out rst,
1
v 5 + 4v 3 + 4v dv = v 4 + x4 + 2v 2 + C
6
v(v 2 + 2)2 dv =
12.
1 3
x
3
+ x + tan1 (x) + C.
18. The hint is that sin(2x) = 2 sin(x) cos(x). Using that,
sin(2x)
dx =
sin(x)
2 sin(x) cos(x)
dx
Review Problems: Chapter 11
1. What does it mean to say that a series converges (Im looking for the denition; be
sure you dene any notation you use).
2. Does the given sequence or series converge or diverge?
1
(a)
n
n=2 n
(b)
(e)
(6)
n1 1n
5
(j)
n=1
(f)
Review Solutions: Chapter 11
1. What does it mean to say that a series converges?
SOLUTION: We dene the nth partial sum sn as follows:
n
S 1 = a1
S 2 = a1 + a2
S3 = a1 + a2 + a3
Sn =
an
k=1
The partial sums Sn form a sequence (of numbers). The (innite) se
Quiz 2 Solutions
Part of this quiz was to see if you could follow the directions!
Solutions are written neatly, clearly and completely using your own paper (up to 10 pts)
Quiz is stapled (up to 10 pts).
Quiz turned in on time (up to 10 pts).
Here are t
Selected Solutions, Section 4.9
10. Note that e2 is a constant, so the antiderivative is e2 C
17. The antiderivative is 2 cos() tan() + C, but notice that the C can change because
sec() has a lot of vertical asymptotes, breaking up the real line. Therefor
/dh+// /0 WI/X/M n3 M 3
J: ._...... ¢ [/)»/D./x/
laul ("+03 /o/)c/ w
an; \ g
5; /((KM / *I m L) m /x/ z /0/x/
44*» (an! a '4
/0"/X/<I =7
To 'hu/ lwcvvcl, (VIM/Mr hv WIS: X3 l 9.
4+ nil/,0;
lo /0/4 n w N r 1
Z % 7 Z ,0 (75) PM) dwux Q,
' , _________
Selected Solutions, Section 5.3
4. This is a good exercise to understand the area function that we described in class.
(a) g(0) = 00 f (t) dt = 0, and g(6) = 06 f (t) dt = 0 by symmetry (it looks like there is
as much positive area as negative.
(b) Estima