Math 181 Fall 2011
Name:
Second 1-Hour Exam
(20 pts) 1. Calculate the arc length of the graph of
9 x2
f (x ) =
over [0, 3].
(20 pts) 2. Determine the limit of the sequence
an =
2n2 + (0.3)n
3n 2 n + 1
(20 pts) 3. Determine whether the improper integral co
Math 181
Fall 2013
Hour Exam 2
11/08/13
Time Limit: 50 Minutes
Name (Print):
This exam contains 9 pages (including this cover page) and 7 problems. Check to see if any pages
are missing. Enter all requested information on the top of this page.
You may not
8.2, #23. The equation of the ellipse is (x/2)2 + (y)2 = 1. The upper half of the ellipse x2 is the graph of f (x) = 1 (x/2)2 . We compute 1 + f (x)2 = 1 + . The 1 (x/2)2 circumference is given by 2 1 3x2 /16 4 dx. 1 x2 /4 0 The integral is improper; the
8.3, #9. A spherical planet of radius R and density has volume V = (4/3)R3 and mass
M = (4/3)R3 . The escape velocity v for an object from this planet is given in Example
5 on page 441: v = 2GM/R. In particular, v = 8GR2 /3 and v is proportional to R
and
8.2, #6. Slice the solid by planes perpendicular to the axis of rotation. The solid is gotten
by rotating the nite region enclosed by the graphs of y = x3 , y = 1, and x = 1. This
region is illustrated below.
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8.1, #1. Break up the rod by slicing it with planes perpendicular to the direction of
the rod. A slice of thickness x at a point x meters from the left end of the rod has
approximately mass (x) dx = (2 + 6x) x g/m. So the rod has total mass
2
2
(2 + 6x) d
7.Rev, #20. Use formula 25 on page 367 to get
5x + 6
5
2x
dx
5
x
dx =
dx + 6
= ln(x2 + 4) + 3 arctan + C.
2+4
2+4
2+4
x
2
x
x
2
2
1
u3 + 3u2 + 3u + 1
1
du = u2 + 3u + 3 ln |u| + C.
2
u
2
u
x dx
1
du
7.Rev, #32. u = x2 , du = 2x dx changes
to
.
2
1 x4
1 u2
0
7.8, #4.
b
ex
dx = ln(1 + ex )
1 + ex
b
1
dx = ln(ln x)
2 x ln x
This diverges as b .
0
b
b
7.8, #17.
b
1
1
dx =
2
ln x
2 x(ln x)
has limit (1/ ln 3) as b .
= ln 2 ln(1 + eb ). This has limit ln 2 as b .
2
= ln(ln b) ln(ln 2) (use the substitution w =
2
7.9, #21. If x 3, then x2 3x and ex e3x . Thus
3
2
ex dx
3
e3x dx =
1
lim e3x
3 b
b
=
3
e9
.
3
2
Similarly if x n, then x2 nx and ex enx . Thus
x2
e
n
dx
nx
e
n
1
dx = lim enx
n b
b
n
2
en
=
.
n
2x2 + 1
2x2
1
behaves like
=
for large x. By the p-test
7.6, #8. LEFT= 0.815997, RIGHT= 0.753497, and TRAP= 0.785724. The actual value
of the integral is /4. Here LEFT is an over estimate and RIGHT is an under estimate
since the integrand 1/(1 + x2 ) is decreasing on [0, 1]. Thus 0.753497 < /4 < 0.815997 and
3
7.7, #16. The statement is true. y 2 1 is concave-up on [0, 1] since the graph of (y 2 1)
1
is a parabola opening upwards. Thus the midpoint rule underestimates 0 (y 2 1) dy.
7.7, #17. The statement is false. The trapezoid rule is exact whenever the graph
7.5, #17. Formula 15 with x = y, p(x) = y 2 , a = 2 gives
1
1
1
y 2 sin 2y dy = y 2 cos 2y + 2y sin 2y + 2 cos 2y + C.
2
4
8
7.5, #18. Formula 11 with x = y, a = 2, b = 7 gives
1
(7 cos 2y sin 7y 2 sin 2y cos 7y) + C.
cos 2y cos 7y dy =
45
7.5, #24. Subst
7.3, #17. Set w = x2 + 4, dw = 2x dx. Then w = 20 if x = 4, w = 5 if x = 1, and
1
x x2 + 4 dx =
4
1
2
5
w1/2 dw =
20
1 3/2
w
3
5
=
20
1 3 /2
(5 203/2 ) = 26.087.
3
7.3, #20. Set w = x2 + 4x + 5, dw = (2x + 4) dx. Then w = 1 if x = 2, w = 5 if x = 0,
and
0
7.4, #9. Set t2 sin t = uv , where u = t2 , v = sin t. Then u = 2t, v = cos t, and
t2 sin t dt = t2 cos t +
2t cos t dt
The second integral is 2t sin t + 2 cos t + C (See example 2 on page 360 of the text). The
nal answer is t2 cos t + 2t sin t + 2 cos t
8.5, #5. Lets say most means something like 65% of the plants. The cumulative
distribution functions of the plants with fertilizer A, with no ferilizer, and with fertilizer
B have value .65 when the heights of the plants are respectively .9, 1.1, and 1.3
8.6, #6. The fraction of the population making between $20000 and $50000 is P (50)
P (20) = 99 75 = 24%. The median income is the x for which P (x) = 50%, that is,
x = $12600. The table for P (x) shows the fastest increase in the range 7.8 x 12.6, so
the
8.Rev, #6. A cut by a plane perpendicular to the x-axis at x meets the doyghnut in an
annulus with outer radius 3 + 1 x2 and inner radius 3 x2 for x in the range
1
2 )2 (3 1 x2 )2 = 12 1 x2 .
1 x 1. The area of this annulus is cfw_(3+ 1 x
1
The volume is
Final Exam, Fall 2013
David Cabrera
Math 181 Calculus II
University of Illinois Chicago
1 / 25
1. Evaluate
x ln(x) dx.
Solution: We use Integration by Parts. Letting u = ln(x) and dv = x dx
1
yields du = x dx and v = 1 x2 . The Integration by Parts formul
Math 181, Final Exam, Fall 2014
Problem 1 Solution
1. (a) Evaluate the integral
ln
x dx.
Solution: We use a property of logarithms to rewrite the integral as
ln
Letting u = ln(x) and dv = dx yields du =
1
x
x dx =
1
2
ln(x) dx.
dx and v = x. Integration b
Math 181, Exam 2, Spring 2013
Problem 1 Solution
1. Compute the following sums
+
(a)
n=3
+
(b)
n=1
2
n(n 1)
2n+3
5 73n2
Solution:
(a) This is a telescoping series. To compute the sum, we decompose the summand as
follows:
2
2
2
=
n(n 1)
n1 n
The N th parti
Math 181, Exam 2, Study Guide 2
Problem 1 Solution
1. Use the trapezoid rule with n = 2 to estimate the arc-length of the curve y = sin x between
x = 0 and x = .
Solution: The arclength is:
b
1+
L=
a
dy
dx
2
dx
1 + (cos x)2 dx
=
0
=
1 + cos2 x dx
0
We now
Math 181, Exam 2, Fall 2014
Problem 1 Solution
1. Integrals, Part I (Trigonometric integrals: 6 points). Evaluate the integral:
sin3 (x) cos(x) dx.
Solution: We begin by rewriting sin3 (x) as
sin3 (x) = sin(x) sin2 (x) = sin(x)(1 cos2 (x).
Then, after usi
Math 181, Exam 1, Fall 2014
Problem 1 Solution
1. Consider the functions
f (x) = 2x2 6x
and g(x) = 6x2 4.
a. Find the x-coordinate of the intersection points of these two graphs.
b. Compute the area of the region bounded by the graphs of the functions f (
10.6, #6. f (x) is an odd function on [, ]. So f (x) cos nx and f (x) sin nx are respectively
odd and even functions on [, ]. In particular,
1
1
a2 =
1
a3 =
a1 =
f (x) cos x dx = 0
f (x) cos 2x dx = 0
f (x) cos 3x dx = 0
and
2
b1 =
2
b2 =
2
b3 =
sin x dx
10.5, #10. Pn (1) approximates cos 1 to four decimal places if the corresponding error En
satises |En | < 0.00005. By the preceding problem we want 1/(n + 1)! < 0.00005. This
happens for n = 7 since 8! = 40320 and 1/8! = 0.000025. For 6 decimal places, we
10.4, #15.
3+
3 3 3
3
+ + + + 10
2 4 8
2
1 1 1
1
= 3 1 + + + + + 10
2 4 8
2
11
3(1 1/2 )
=
1 1/2
3(211 1)
=
.
210
10.4, #23. (a) The height to which the ball rises after it hits the oor for the n-th time is
10(3/4)n feet.
(b) The ball has traveled a total
10.2, #12. If f (x) = x-1 , then f (x) = -x-2 , f (x) = +2!x-3 , and, in general, f (n) (x) = (-1)n n!x-(n+1) . Evaluating these derivatives at 1 we have f (1) = 1, f (1) = -1, f (1) = 2!, and f (n) (1) = (-1)n n!. The taylor series is 1 - (x - 1) + (x -
10.3, #11. We multiple the two series:
1
1
1
1
et cos t = 1 + t + t2 + t3 + (1 t2 + t4 +
2
6
2
24
1 2 1 2
1 3 1 3
1
1
1
= 1 + t + ( t t ) + ( t t ) + ( t4 t4 + t4 ) +
2
2
6
2
24
4
24
1 3 1 4
= 1 + t t t + .
3
6
10.3, #15. We expand the three functions i
10.4, #1. This is a geometric series with a = 2 and r = 1/2, its sum is 4.
10.4, #2. This is a geometric series with a = 1 and r = 1/2, its sum is 2/3.
10.4, #3. This is not a geometric series: the (n + 1)-st term comes from the n-th term by
multiplying b