Math 1210-001
Final Exam Review
1 August 2013
This exam consists of n problems and 1 bonus problem, and will last 120
minutes for some integer n. Answer the questions in the spaces provided.
You may use a scientific or graphing calculator. You may not have any books,
notes, cell phones etc with you at your desk. You must show all of your work
to receive credit. Anyone found violating these policies will be asked to leave
the exam immediately, and will receive a 0 on the exam. Good luck!
For review exercises concerning sections 1.1 through 3.4, see the review materials for
exams 1 and 2.
1. Graph the function f (x) which has the following characteristics:
• x-intercepts are −2 and 2 and y-intercept is 16.
• no vertical asymptotes
• f (x) is positive on (−∞, −2), (−2, 2), and (2, ∞) and is never negative
• lim f (x) = ∞ and lim f (x) = ∞
x→−∞
x→∞
• decreasing on (−∞, −2), and (0, 2) and increasing on (−2, 0) and on (2, ∞)
• two local minimum points: (−2, 0) and (2, 0) and one local maximum point: (0, 16).
• no singular points
• concave up on (−∞, −1.1547) and (1.1547, ∞)
• concave down on (−1.1547, 1.1547)
• two points of inflection: (−1.1547, 7.1111) and (1.1547, 7.1111)
Solution: This is the function f (x) = (x − 2)2 (x + 2)2 . Check your graph using
your calculator or WolframAlpha.com.
2. Graph the function f (x) which has the following characteristics:
• x-intercepts are −2 and 2 and y-intercept is −8.
• no vertical asymptotes
• f (x) is negative on (−∞, −2), (−2, 2), and (2, ∞) and is never positive
• lim f (x) = −∞ and lim f (x) = −∞
x→−∞
x→∞
• increasing on (−∞, −2), and (0, 2) and decreasing on (−2, 0) and on (2, ∞)
• two local maximum points: (−2, 0) and (2, 0) and one local minimum point: (0, −8).
• no singular points
• concave down on (−∞, −1.1547) and (1.1547, ∞)
• concave up on (−1.1547, 1.1547)
• two points of inflection: (−1.1547, −3.5555) and (1.1547, −3.5555)
Solution: This is the function f (x) = − 12 (x − 2)2 (x + 2)2 . Check your graph using
a graphing calculator or WolframAlpha.com.
3. Graph the function f (x) which has the following characteristics:
• x-intercepts are −2 and 2 and y-intercept is 8.
• no vertical asymptotes
• f (x) is negative on (−∞, −2), and positive on (−2, 2) and (2, ∞)
• lim f (x) = −∞ and lim f (x) = ∞
x→−∞
x→∞
• increasing on (−∞, − 23 ) and (2, ∞), and decreasing on (− 32 , 2)
• one local maximum point: (− 23 , 9.48148) and one local minimum point: (2, 0)
• no singular points
• concave down on (−∞, 32 ) and concave up on ( 23 , ∞)
• one point of inflection: ( 23 , 4.7407)
Solution: This is the function f (x) = (x − 2)2 (x + 2). Check your graph using a
graphing calculator or WolframAlpha.com.
4. Graph the function f (x) which has the following characteristics:
• no x-intercepts, y-intercept is − 14
• vertical asymptotes to x = 2 and x = −2
• f (x) is negative on (−2, 2), and positive on (−∞, −2) and (2, ∞)
• lim f (x) = 0 and lim f (x) = 0
x→−∞
x→∞
• increasing on (−∞, −2) and (−2, 0), and decreasing on (0, 2) and (2, ∞)
• one local maximum point: (0, − 14 ) and no local minimum points
• no singular points other than x = −2 and x = 2 (vertical asymptotes)
• concave down on (−2, 2) and concave up on (−∞, −2) and (2, ∞)
• no points of inflection
Page 2
1
. Check your graph using a graphSolution: This is the function f (x) = − (x−2)(x+2)
ing calculator or WolframAlpha.com.
5. Let f (x) = 3x3 − 3x2 + 3x − 3. Find all numbers c which satisfy the conclusion of the
Mean Value Theorem for derivatives for f (x) on the interval [−1, 2].
Solution: -0.548584, 1.21525
6. Exhibit the first three steps of the Bisection Method for finding the root of f (x) = x2 − 3
which falls between 1 and 2. In particular, find a1 , b1 , a2 , b2 , a3 , and b3 , as well as the
final estimate for the root and the error of the final estimate.
Solution: First, notice that f (1) = −2 and f (2) = 1.
a1 = 1 b1 = 2 m1 = 1.5 f (m1 ) = −0.75
a2 = 1.5 b2 = 2 m2 = 1.75 f (m2 ) = 0.0625
a3 = 1.5 b3 = 1.75 m3 = 1.625 f (1.625) = −0.3594
Acceptable answers for final estimate and error, depending on whether f (m3 ) is
calculated or not:
r ≈ 1.625, E = E3 = (.25)/2 = .125
r ≈ 1.6875, E = E4 = .125/2 = .0625
7. Find an antiderivative of the function g(x) = 3x2 + 4x + 17.
Solution: There are many antiderivatives of this function.
One is G(x) = x3 + 2x2 + 17x.
8. Find the general antiderivative of the function h(x) =
6
Solution: H(x) = 56 x 5 + 2x + C
Page 3
√
5
x + 2.
9. Calculate the Riemann Sum for the function g(x) = x3 on the interval [3, 5] with the
following partition into 6 subintervals:
3 < 3.5 < 3.9 < 4.1 < 4.75 < 4.8 < 5
x̄1 = 3.5, x̄2 = 3.75, x̄3 = 4.0, x̄4 = 4.1, x̄5 = 4.75, x̄6 = 4.9
(Round your answer to 5 decimal places.)
Solution:
RP = (3.5 − 3)g(3.5) + (3.9 − 3.5)g(3.75) + (4.1 − 3.9)g(4.0)
+ (4.75 − 4.1)g(4.1) + (4.8 − 4.75)g(4.75) + (5 − 4.8)g(4.9)
= 129.01829
√
10. Set up the Riemann Sum for the function f (x) = x + 1 on the interval [1, 6], using a
regular partition of 10 subintervals with the sample point xi being the midpoint of the
ith interval. Write your answer in sigma notation in terms of i and constants. You do
not need to evaluate the sum.
Solution: First, notice that xi = 1 +
xi = xi−12+xi = 1 + 2i−1
.
4
RP =
10
X
5
i
10
= 1 + 2i , ∆xi =
5
10
f (xi ) · ∆xi
2i − 1
1
=
f 1+
4
2
i=1
"
#
r
10
1X
2i − 1
=
1+
+1
2 i=1
4
R0
1
x t
dt.
Solution: F 0 (x) = − x1
12. Evaluate the indefinite integral
R
3
2
2x − 3x +
Page 4
1
2
and
(1)
i=1
10 X
11. Find F 0 (x) if F (x) =
=
1
√
3x
dx.
(2)
(3)
Solution:
Z
x4
3 2
1
dx
=
− x3 + x /3 + C
2x3 − 3x2 + √
3
2
2
x
13. Evaluate the indefinite integral
Solution:
Z
x cos(x2 − 2) sin5 (x2 − 2) dx.
R
x cos(x2 − 2) sin5 (x2 − 2) dx =
R1 14. Evaluate the definite integral
−1
2x3 − 3x2 +
1
√
3x
1
sin6 (x2 − 2) + C
12
dx
Solution:
Z
1
1
2x3 − 3x2 + √
dx
3
x
−1
1
4
3 2/3
x
3
−x + x
=
2
2
−1
= −2
R − √2
√
15. Evaluate the definite integral
2
(4)
(5)
(6)
x cos(x2 − 2) sin5 (x2 − 2) dx
Solution:
Z
√
− 2
√
x cos(x2 − 2) sin5 (x2 − 2) dx
(7)
2
− √ 2
1
=
sin6 (x2 − 2) √
12
2
(8)
** For more practice with definite and indefinite integrals, see the exercises §4.4. Answers
to odd exercises are in the back of the book.
√
16. Find the average value of f (x) = x3 sin(5x4 ) on the interval [0, 4 π].
Page 5
Solution:
Average Value of f =
=
=
=
Z √
4π
1
√
x3 sin(5x4 ) dx
4
π−0 0
4π
√
1
1
4
√
− cos(5x )
4
20
π
0
1
√ [− cos(5π) + cos(0)]
20 4 π
1
√
10 4 π
(9)
(10)
(11)
(12)
17. Find all values of c that satisfy the Mean Value Theorem for integrals for f (x) = x2 on
the interval [−4, 4].
Solution: c =
q
16
3
and c = −
q
16
3
18. Find the area of the region bounded by y = x − x3 and the x-axis.
Solution:
1
2
19. Find the area bounded by y = x2 and y = 4.
Solution:
32
3
20. The base of a solid is the region bounded by y = x2 and y = x. Every cross section in a
plane perpendicular to the x-axis is a square, one side of which is in the base. Find the
volume of the solid.
Solution:
1
30
21. The region bounded by y = x2 , x = 3 and y = 0 is rotated around the y-axis. Find the
volume of the resulting solid.
Page 6
Solution:
81π
2
√
22. The region bounded by y = x, x = 9, and the x-axis is rotated around the y-axis.
Find the volume of the resulting solid.
Solution:
972π
5
23. The region bounded by y = x2 and y = x + 2 is revolved about the line x = −1. Use
shells to find the volume.
Solution:
27π
2
24. Write an integral whose value is the length of the parametric curve {x = t, y = t2 } from
t = 1 to t = 4.
Z
4
Solution:
√
1 + 4x2 dx
1
25. Write an integral whose value is the length of the curve x = 2y − y 3 from y = 1 to y = 6.
Z
Solution:
6
p
1 + (2 − 3y 2 )2 dy
1
Page 7