math 10b review sheet for exam 1 {...
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Math 10b Review Sheet for Exam 1 – SOLUTIONS
• The first Math 10b midterm will be Monday, February 12th, 7:30 – 9:30 p.m.
• Location: Gerstenzang 123
• The exam will cover:
– Section 3.6: Inverse Trig
– Appendix F: Sigma Notation
– Section 5.1: Areas and Distances
– Section 5.2: The Definite Integral
– Section 5.4: Fundamental Theorem of Calculus (Part I)
• No calculators will be allowed on the exam.
Our exams are closed book: no notes, books, calculators, cell phones or internet-connected devices will be allowed.
• On the exam, you will be asked to show all your work.
We are grading your ability to clearly communicate your understanding of how to solveeach problem much more than your ability to get the correct answer. In fact, a correctanswer with little or no work might not receive any points at all.
Showing your work clearly gives us justification for assigning partial credit.
• We will have at least one “explain what you did” question.
As you study, make sure you are able to explain your work. Practice by explaining toyour friends, your instructor (in office hours), or tutors (in evening help).
• On the exam, you will be given the following three summation formulae (so you do*not* need to memorize these):
n∑i=1
i =n(n+ 1)
2
n∑i=1
i2 =n(n+ 1)(2n+ 1)
6
n∑i=1
i3 =
[n(n+ 1)
2
]2
OVER for Practice Exam 1 SOLUTIONS→
Practice Exam 1 – SOLUTIONS (Problems 1 - 10)
NOTE: You may use the following three summation formulae whenever needed. Theseformulae will also be given to you on the exam.
(1)n∑i=1
i =n(n+ 1)
2, (2)
n∑i=1
i2 =n(n+ 1)(2n+ 1)
6, (3)
n∑i=1
i3 =
[n(n+ 1)
2
]2
1. Below is the graph of f(x).
1 2 3 4 5 6 7
−3
−2
−1
0
1
2
3
(a) Find
∫ 7
0
f(x)dx.
(Note that the part of the graph between x = 5 and x = 7 is a semi-circle.)
Solution:∫ 7
0
f(x)dx =
∫ 1
0
f(x)dx+
∫ 2
1
f(x)dx+
∫ 3
2
f(x)dx+
∫ 5
3
f(x)dx+
∫ 7
5
f(x)dx
=1
2(1)(2)− 1
2(1)(2)− (1)(2)− 1
2(2)(2) +
π(1)2
2
=π
2− 4
(b) Using your answer from part 1a, find
∫ 7
0
f(x) + 2
5dx.
Solution: ∫ 7
0
f(x) + 2
5dx =
1
5
∫ 7
0
f(x) dx+
∫ 7
0
2
5dx
=1
5(π
2− 4) +
2
5(7− 0)
=π
10+ 2
2. Compute the following:
(a) sin−1(−√
32
)Solution: Since sin
(−π3
)= −
√3
2, we have sin−1
(−√
32
)= −π
3.
(b) arctan(tan(
7π4
))Solution: arctan
(tan(
7π4
))= arctan(−1) = −π
4
(c) cos(arcsin
(513
))Solution: Let θ = arcsin
(513
). Then sin θ = 5
13.
513
aθ
a2 + 52 = 132
a2 = 169− 25
a =√
144
a = 12
So cos(arcsin
(513
))= cos θ = 12
13
3. Find the derivatives of the following functions:
(a) f(x) = ln(arctan(2x))
Solution: f ′(x) =
(1
arctan(2x)
)(1
1 + 4x2
)(2)
(b) h(x) =
∫ x2
x
sin(t2) dt
Solution:
h(x) =
∫ x2
x
sin(t2) dt
=
∫ 0
x
sin(t2) dt+
∫ x2
0
sin(t2) dt
= −∫ x
0
sin(t2) dt+
∫ x2
0
sin(t2) dt
Since f(t) = sin(t2) is continuous on (−∞,∞), we can use part I of the Fundamen-tal Theorem of Calculus (FTC I). By FTCI, we know d
dx
(∫ x0
sin(t2) dt)
= sin(x2).For the second integral, we use the chain rule together with FTC I (where our “in-
side function” is g(x) = x2, so g′(x) = 2x) and so ddx
(∫ x20
sin(t2) dt)
= sin(x4)2x.
Soh′(x) = − sin(x2) + 2x sin(x4)
4. Evaluate the following:24∑
k=−2
(4k − 3). (Your answer should be a number.)
Solution 1: Let i = k + 3, so k = i− 3. Then
24∑k=−2
(4k − 3) =27∑i=1
(4(i− 3)− 3)
=27∑i=1
4i− 15
= 427∑i=1
i−27∑i=1
15
= 4
(27(27 + 1)
2
)− 27(15)
= 4(27)(14)− 27(15)
= 27(56− 15)
= 27(41)
= 1107
Solution 2: Here’s another way you can do the problem:
24∑k=−2
(4k − 3) = (4(−2)− 3) + (4(−1)− 3) + (4(0)− 3) +24∑k=1
(4k − 3)
= −11− 7− 3 + 424∑k=1
k −24∑k=1
3
= −21 + 424(25)
2− 3(24)
= −21 + 24(50− 3)
= −21 + 24(47)
= 1107
Note: there are some other variations as well.
5. Consider the definite integral
∫ 3
0
2−x + 1 dx.
(a) Write out the sigma notation (with xi, etc.) for the left-hand Riemann sum L3
for this integral. Then find ∆x and xi and finally plug them in to compute L3.
Solution: L3 =2∑i=0
f(xi) ∆x = f(x0)∆x+ f(x1)∆x+ f(x2)∆x
∆x =a− bn
=3
3= 1 xi = a+ i∆x = 0 + i(1) = i
L3 = (20 + 1)(1) + (2−1 + 1)(1) + (2−2 + 1)(1) = 4 +1
2+
1
4=
19
4
(b) On the axes below, carefully sketch the graph of f(x) = 2−x + 1, shading in thearea represented by the above integral. Then draw in the rectangles that represent
the area you computed in 5a.
0 1 2 3
1
2
−1
x
y
(c) Is your estimate in 5a an overestimate or an underestimate of
∫ 3
0
2−x + 1 dx, or
is there not enough information to determine one way or the other?
Briefly explain your answer.
Our estimate is an overestimate since the function is decreasing and we are usingleft-hand end-points for the height of the rectangles (so the y-values of the functionover the remainder of each subinterval are less than the y-value we’re using forthe height of the rectangle for each subinterval).
6. We will compute the volume of a sphere using calculus!
z1
√1− z2
The volume of this solid sphere can befound by taking the following integral:
∫ 1
−1
π(√
1− z2)2
dz
Compute this integral using the limit defi-nition of the integral to find the volume ofthe sphere of radius one!
Solution: ∆z =b− an
=1− (−1)
n=
2
nzi = a+ i∆z = −1 +
2i
n
∫ 1
−1
π(√
1− z2)2
dz = π
∫ 1
−1
1− z2 dz
= π limn→∞
n∑i=1
[1−
(−1 +
2i
n
)2]
2
n
= π limn→∞
2
n
n∑i=1
1−(
1− 4i
n+
4i2
n2
)= π lim
n→∞
2
n
n∑i=1
(4i
n− 4i2
n2
)
= π limn→∞
(8
n2
n∑i=1
i− 8
n3
n∑i=1
i2
)
= π limn→∞
[8
n2
(n(n+ 1)
2
)− 8
n3
(n(n+ 1)(2n+ 1)
6
)]= π lim
n→∞
[4
(n+ 1
n
)− 4
3
(2n2 + 3n+ 1
n2
)]= π
(4− 8
3
)=
4π
3
7. Write the following sum in sigma notation with index starting at 0.
1− 3 + 5− 7 + 9− 11 + 13− 15
Answer:7∑
n=0
(−1)n(2n+ 1)
8. Cameron is training for a marathon. Suppose the function v(t) gives Cameron’s velocityat time t, where velocity is given in miles per hour and the time is given in t hourssince Cameron left home. (Cameron’s training route is down a straight road.)
(a) Suppose we know the following information about v(t).
t 0 12
1 32
2
v(t) 3 6 5 −5 −4
Use a right-hand Riemann sum with four intervals to estimate
∫ 2
0
v(t) dt.
Solution: ∆t =b− an
=2− 0
4=
1
2ti = a+ i∆t = 0 +
i
2=i
2∫ 2
0
v(t) dt ≈ R4 =4∑i=1
v(ti)∆t
= 6
(1
2
)+ 5
(1
2
)− 5
(1
2
)− 4
(1
2
)= 1 mile
(b) What does
∫ 2
0
v(t) dt represent in terms of the scenario described above? (De-
scribe in words.)
Answer: This integral represents how far Cameron is from home after runningfor two hours.
(c) What does
∫ 2
0
|v(t)| dt represent in terms of the scenario described above? (De-
scribe in words; please do not compute it.)
Answer: This integral represents the total distance Cameron traveled after run-ning for two hours.
9. Find the x-values of the local extrema for the function
f(x) =
∫ x
0
t2 + 3t+ 2 dt.
Identify each as a local maximum or a local minimum.
Solution: Polynomials are continuous, so by FTCI, f ′(x) = x2 + 3x+ 2.
f ′(x) = 0
x2 + 3x+ 2 = 0
(x+ 2)(x+ 1) = 0
So the critical numbers are x = −2 and x = −1.
f ′′(x) = 2x+ 3
f ′′(−2) = −1 < 0, so x = −2 is a local maximum
f ′′(−1) = 1 > 0, so x = −1 is a local minimum
10. Below are three “true or false” questions. For each, determine whether the statementis true or false.1 If a statement is true, briefly explain why. If a statement is false,give an example (a function or a formula or a graph) that shows that it’s false.
Let f(x) be a continuous function on the real numbers, and let
g(x) =
∫ x
0
f(t)dt
(a) True or false: If f is increasing at x = 2 then g(2) must be positive.
FALSE:
0 1 2 3
1
−1
−2
f
This function f is increasing at x = 2, but g(2) is the negative of the area betweenthe x-axis and f , and so g(2) is negative.
(b) True or false: If f ′(2) > 0 then g is increasing at x = 2
FALSE: The same example for part 10a will work here. Since f is increasingat x = 2, we have f ′(2) > 0, but as x increases from x = 2, the function g isdecreasing since f is negative.
(c) True or false: If f(x) ≤ 0 for all x, then g(x) ≤ 0 for all x.
FALSE: Take f(x) = −1. Then f is negative on (−∞,∞). But
g(−3) =
∫ −3
0
−1 dx = −∫ 0
−3
−1 dx = −(−1)(0− (−3)) = 3,
so g(−3) > 0.
OVER for Practice Exam 2 SOLUTIONS →1When we say a statement is “true” we mean it must always be true. When we say a statement is “false”
we mean it could possibly be false.
Practice Exam2 – SOLUTIONS (Problems 11 - 24)
NOTE: You may use the following three summation formulae whenever needed. Theseformulae will also be given to you on the exam.
(1)n∑i=1
i =n(n+ 1)
2, (2)
n∑i=1
i2 =n(n+ 1)(2n+ 1)
6, (3)
n∑i=1
i3 =
[n(n+ 1)
2
]2
11. Evaluate the following:
(a) sin(arctan( 1√3)) = sin(π
6) = 1
2
(b) arcsin(sin 2π) = sin−1(0) = 0
12. Find the domain of f(x) = ln(arcsin(x)).
The domain of arcsin(x) is [−1, 1]. Since f(x) = ln(arcsin(x)), we must have arcsin(x) >0, so x > 0. Thus the domain of f(x) is (0, 1].
13. Compute the following limit
limh→0
tan−1(3 + h)− tan−1(3)
h.
Let f(x) = tan−1(x). Then limh→0
tan−1(3 + h)− tan−1(3)
h= f ′(3) =
1
1 + 32=
1
10.
14. Find the equation of the line tangent to the graph of f(x) = arctan(ex) at x = 0.
f(x) = arctan(ex) ⇒ f ′(x) =1
1 + (ex)2 · ex. So the slope of the tangent line is
f ′(0) = 12· 1 = 1
2. The point of tangency is (0, f(0)) = (0, π
4). So the equation of the
tangent line is y − π4
= 12(x− 0), which can also be written as y = 1
2x+ π
4.
15. One or more of these functions is an antiderivative of f(x) =x
1 + x2 . Which one(s)?
F (x) is an antiderivative of f(x) if F ′(x) = f(x). So differentiate each of the four
functions and see which has derivative f(x) =x
1 + x2 .
(a)d
dx
(12
ln(1 + x2))
=1
2· 1
1 + x2 · 2x =x
1 + x2 .
(b)d
dx
( x2
2+ lnx
)= x+
1
x=
x2 + 1
x.
(c)d
dx
(x tan−1 x
)= 1 · tan−1 x+ x · 1
1 + x2 = tan−1 x+x
1 + x2 .
(d)d
dx
(tan−1(x2)
)=
1
1 + (x2)2 · 2x =2x
1 + x4 .
So F (x) = 12
ln(1 + x2) is an antiderivative of f(x), and the other functions are not.
16. Write the following sum using sigma notation. (There is more than one correct answer.)
1
1 · 2−
1
2 · 3+
1
3 · 4−
1
4 · 5+ · · ·+
1
49 · 50.
One correct answer is49∑i=1
(−1)i+11
i(i+ 1).
Another correct answer is48∑i=0
(−1)i1
(i+ 1)(i+ 2).
Yet another is50∑i=2
(−1)i1
(i− 1)(i).
17. Let f(x) = 4− x2. Estimate the area under the graph of f(x) over [0, 2] by finding
(a) the Riemann sum R4. Is this an overestimate or underestimate?
(b) the Riemann sum L4. Is this an overestimate or underestimate?
(c) Find the exact area by computing limn→∞
Rn.
f(x) = 4− x2. The four subintervals are [0, 12], [1
2, 1], [1, 3
2], [3
2, 2]. Note that ∆x = 1
2.
(a)
R4 =4∑i=1
f(xi) ∆x
= f(1/2)∆x+ f(1)∆x+ f(3/2)∆x+ f(2)∆x
=15
4· 1
2+ 3 · 1
2+
7
4· 1
2+ 0 · 1
2
=17
4.
This is an underestimate.
(b)
L4 =4∑i=1
f(xi−1) ∆x
= f(0)∆x+ f(1/2)∆x+ f(1)∆x+ f(3/2)∆x
= 4 · 1
2+
15
4· 1
2+ 3 · 1
2+
7
4· 1
2
=25
4.
This is an overestimate.
(c) Since a = 0 and b = 2, ∆x = 2−0n
= 2n. xi = 0 + i · ∆x = 2i
n. So we get the
Riemann sumn∑i=1
(4− (2i
n)2)
2n. Simplifying gives
n∑i=1
(4− (2i
n)2)
2n
= 2n
n∑i=1
(4− 4i2
n2
)= 2
n
n∑i=1
4− 2n
n∑i=1
4i2
n2 = 8n
n∑i=1
1− 8n3
n∑i=1
i2
= 8n· n− 8
n3 · n(n+1)(2n+1)6
= 8− 8n3+12n2+4n3n3 .
Then the exact area is limn→∞
(8− 8n3+12n2+4n
3n3
)= 8− 8
3= 16
3.
18. Let f(x) ≥ 0 on the interval [0, 5]. Suppose that we estimate the area under the graphof y = f(x) using R5 and get R5 = 8.
Now suppose that you want to estimate the the area under the graph of the functiong(x) = 3 + 2f(x) on the interval [0, 5]. What is R5 for g(x)?
Solution:
For f(x), R5 = f(1) ·∆x+ f(2)∆x+ f(3)∆x+ f(4)∆x+ f(5)∆x = 8. Since ∆x = 1,we can write this as follows: f(1) + f(2) + f(3) + f(4) + f(5) = 8.
For g(x), R5 = g(1)∆x+ g(2)∆x+ g(3)∆x+ g(4)∆x+ g(5)∆x. Since g(x) = 3 + 2f(x)and ∆x = 1, we can rewrite this sum as[
3 + 2f(1)]
+[3 + 2f(2)
]+[3 + 2f(3)
]+[3 + 2f(4)
]+[3 + 2f(5)
]=[3 + 3 + 3 + 3 + 3
]+[2f(1) + 2f(2) + 2f(3) + 2f(4) + 2f(5)
]= 3 · 5 + 2
[f(1) + f(2) + f(3) + f(4) + f(5)
].
We know that the sum in brackets equals 8, so the entire expression equals 15+ 2(8) =31. So R5 for g(x) equal 31.
19. Express the following limit as a definite integral. Do not evaluate the integral. Notethat there are many correct answers.
limn→∞
n∑i=1
sin2(π + πi
2n
)π2n.
We have to choose a value of a. In this problem it looks like ∆x = π2n
. So π + πi2n
=π+ i∆x is good choice for xi, which would make a = π. (Note: there are other choicesfor a, but this is the most obvious.) Since ∆x = b−a
nand a = π, we get that b−π = π
2,
so b = 3π2
. Finally, the function in the integrand is sin2(π + πi
n
)so f(xi) = sin2(xi),
so f(x) = sin2 x. Hence the integral is∫ 3π/2
π
sin2 x dx.
20. Is the following statement true or false? Explain your answer.
The integral
∫ 3
0
(x2 − 4) dx represents the area of the region between the
graph of f(x) = x2 − 4 and the x-axis over the interval [0, 3].
False.
∫ 3
0
(x2 − 4) dx = −3, which is clearly not the area of the region since area
must be positive. Instead,
∫ 3
0
(x2−4) dx represents the signed or net area between the
graph of f(x) = x2 − 4 and the x-axis over the interval [0, 3]. In other words,∫ 3
0
(x2 − 4) dx = −(
area between the graph of f(x) and the x-axis over (0, 2))
+(
area between the graph of f(x) and the x-axis over (2, 3)).
21. Let f and g be continuous functions. Suppose we know that∫ 1
0
f(x) dx = 1;
∫ 2
1
f(x) dx = −3;
∫ 1
0
g(x) dx =1
2;
∫ 2
1
g(x) dx = 2 and
∫ 3
0
g(x) dx = 5.
Some of the integrals listed below can be evaluated using this information along withproperties of integrals and others can’t. Find the ones that can, and evaluate them.
(a)
∫ 1
0
(f(x) + g(x)) dx =
∫ 1
0
f(x) dx+
∫ 1
0
g(x) dx = 1 + 12
= 32.
(b)
∫ 2
0
f(x) dx =
∫ 1
0
f(x) dx+
∫ 2
1
f(x) dx = 1 +−3 = −2.
(c)
∫ 1
0
f(x)g(x) dx can’t be computed from given information.
(d)
∫ 3
1
g(x) dx =
∫ 3
0
g(x) dx−∫ 1
0
g(x) dx = 5− 12
= 92.
(e)
∫ 2
1
(5f(x)+1) dx = 5
∫ 2
1
f(x) dx+
∫ 2
1
1 dx = 5(−3)+1(2−1) = 5(−3)+1 = −14.
(f)
∫ 2
1
1
g(x)dx can’t be computed from given information.
(g)
∫ 3
2
f(x− 1) dx =
∫ 2
1
f(x) dx since the graph of y = f(x− 1) is simply the graph
of y = f(x) shifted to the right by one unit. So
∫ 3
2
f(x− 1) dx = −3.
22. Let F (x) =
∫ x
−2
f(t) dt, where f(x) =
x if x ≤ 0
2x if 0 ≤ x < 1
2 if 1 ≤ x < 3
5− x if x ≥ 3
.
(a) Sketch the graph of f(x).
y x x 0<if( )=
y 2x 0 x 1< <if( )=
y 2 1 x 3< <if( )=
y 2 x 4!( )!( ) 3 x 5< <if( )=
y x! 5+ x 3>if( )=
(b) Use the graph to evaluate F (0), F (2), F (4) and F (6).
Remember that F (x) is the net (i.e., signed) area under the graph of f(t) over[−2, x]. Computing net areas under the graph of f(x), we get
• F (0) =
∫ 0
−2
f(t) dt = 12(2)(−2) = −2
• F (2) =
∫ 2
−2
f(t) dt = −2 + 12(1)(2) + (1)(2) = 1
• F (4) =
∫ 4
−2
f(t) dt = 1 + (1)(2) + 12(1)(2 + 1) = 4.5
• F (6) =
∫ 6
−2
f(t) dt = 4.5 + 12(1)(1)− 1
2(1)(1) = 4.5
(c) Find F ′(2) and F ′′(2).
F ′(2) = f(2) = 2. F ′′(2) = f ′(2) = 0.
(d) At what point(s) in the interval [−2, 6] does F (x) = 0?
F (x) = 0 when x = −2 and when x = 1.5.
(e) At what point in [−2, 6] does F (x) achieve its absolute (i.e., global) maximumvalue?
F (x) achieves its global max in [−2, 6] at x = 5.
(f) At what point in [−2, 6] does F (x) achieve its absolute (i.e., global) minimumvalue?
F (x) achieves its global min in [−2, 6] at x = 0.
(g) Is F (x) concave up or down at x = 4? Why?
F (x) is concave down at x = 4 since F ′(x) is decreasing at x = 4.
23. We cut a circular disk of radius r into n circular sectors, as shown in the figure, bymarking the angles θi at which we make the cuts (θ0 = θn can be considered to bethe angle 0). A circular sector between two angles θi and θi+1 has area 1
2r2 ∆θ where
∆θ = θi+1 − θi.
���
������
���������
We let An =n−1∑i=0
1
2r2 ∆θ. Then the area of the disk, A, is given by:
(a) An, independent of how many sectors we cut the disk into.
(b) limn→∞
An
(c)∫ 2π
012r2dθ
(d) all of the above
Answer: (d). Answers (b) and (c) are equivalent. However, limn→∞
An = An for all n.
24. Sketch the graph of an increasing function f(x) such that f ′(x) and F (x) =
∫ x
0
f(t) dt
are both decreasing.
Since f ′(x) is decreasing, f(x) must be concave down. Since F (x) =
∫ x
0
f(t) dt is
decreasing, its derivative must be negative, so f(x) must be negative. So for x ≥ 0,f(x) must be negative, increasing and concave down. Here’s one possible graph:
y6
x 1+! x 0>if" #$ %=
OVER for Practice Problems SOLUTIONS →
Practice Problems – SOLUTIONS (Problems 25 - 50)
NOTE: You may use the following three summation formulae whenever needed. Theseformulae will also be given to you on the exam.
(1)n∑i=1
i =n(n+ 1)
2, (2)
n∑i=1
i2 =n(n+ 1)(2n+ 1)
6, (3)
n∑i=1
i3 =
[n(n+ 1)
2
]2
25. In each of the following, find f ′(x).
(a) f(x) = x2esin−1(3x)
f ′(x) = 2xesin−1(3x) + x2esin−1(3x) · 1√1− (3x)2
· 3
(b) f(x) = ln(√
arctanx)
f ′(x) =1√
arctanx· 1
2(arctanx)−
12 · 1
1 + x2
26. Find the general antiderivative of each of the following functions f(x):
(a) f(x) = 3x +1
4· 1
1 + x2 , so F (x) =3x
ln 3+
1
4tan−1 x+ C.
(b) f(x) = sec x tanx− 2 · 1√1− x2
, so F (x) = sec x− 2 sin−1 x+ C.
27. Find the following limits:
(a) limx→0
tan−1(5x)
ex − e−x is an indeterminate form of type0
0, so use l’Hopital’s rule:
limx→0
tan−1(5x)
ex − e−x = limx→0
5
1 + 25x2
ex + e−x=
5
2.
(b) limx→−∞
2 tan−1 x = 2(
limx→−∞
tan−1 x)
= 2
(− π
2
)= −π.
28. Do problems 11 and 16 on pages 341–2 of the text. In problem 16, use right handendpoints for your estimate.
Problem 11: The subintervals are [0, .5], [.5, 1], [1, 1.5], [1.5, 2], [2, 2.5], [2.5, 3], and∆x = .5.
(a) The speed is increasing over [0, 3] so use left endpoints to get a lower estimate:(0)(.5) + (6.2)(.5) + (10.8)(.5) + (14.9)(.5) + (18.1)(.5) + (19.4)(.5) = 34.7 feet.
(b) The speed is increasing over [0, 3], so use right endpoints to get an upper estimate:(6.2)(.5) + (10.8)(.5) + (14.9)(.5) + (18.1)(.5) + (19.4)(.5) + (20.2)(.5) = 44.8 feet.
Problem 16: Use 6 rectangles. The car’s velocity is measured in km/hr, so convertthe units on the t-axis from seconds to hours. So ∆t = 5
3600= 1
720. Then the distance
is approximately
50 · 1720
+ 80 · 1720
+ 95 · 1720
+ 110 · 1720
+ 115 · 1720
+ 120 · 1720
= 570720≈ .79167 km
Note: The height of each rectangle is an estimate. Your estimates may differ slightlyfrom those given above.
29. Let f(x) be continous on [a, b]. Consider the Riemann sum Rn =n∑i=1
f(xi)·∆x.
(a) Explain why ∆x equals b−an
.
b−a is the length of the interval [a, b]. We divide [a, b] into n subintervals of equallength, so each subinterval has length b−a
n, so ∆x = b−a
n.
(b) Explain why xi equals a+ i ·∆x.
We divide the interval [a, b] into n subintervals of width ∆x. (See picture below.)
a b
x0 x1 x2 x3 · · · xi−1 xi · · · xn−1 xn︸ ︷︷ ︸∆x
Since xi is the right endpoint of the ith subinterval,
xi = a+ ∆x+ ∆x+ · · ·+ ∆x︸ ︷︷ ︸i times
= a+ i ·∆x.
(c) Suppose f(x) ≥ 0 on [a, b]. What does the expression f(xi)·∆x represent?
If f(x) ≥ 0 on [a, b], then f(xi) ·∆x represents the area of the ith rectangle (usingright endpoints for the heights of the rectangles). The reason: the base of the ithrectangle is the subinterval [xi−1, xi], which has length ∆x. The height of the ithrectangle is f(the right endpoint) = f(xi).
y 2 0.1x2
!=
x 2 0 y 2 0.4!< <if( )=
x 1 0 y 2 0.4!< <if( )=
y 2 0.1 2( ) 2! 0.99 x 2< <if( )=
0 y 2 0.1 2( ) 2! 0.99 x 2< <if( )< <
x 0=
y 0=
2
1.6
xi−1 xi
(xi, f(xi))
f(x)
(d) Suppose f(x) ≤ 0 on [a, b]. What does the expression f(xi)·∆x represent?
If f(x) ≤ 0 on [a, b], then f(xi) ·∆x represents the the opposite of the area of theith rectangle (using right endpoints for the heights of the rectangles).
y 0.1x22!=
x 2 0.1 2( ) 2 2! y 0< <if( )=
x 1 0.1 2( ) 2 2! y 0< <if( )=
y 0.1 2( ) 2 2! 0.99 x 2< <if( )=
0.1 2( ) 2 2! y 0 0.99 x 2< <if( )< <
x 0=
y 0=
xi−1 xi
(xi, f(xi))
f(x)
30. Let f(r) be the function drawn below. Estimate
∫ 3
−3
f(r) dr by computing the Riemann
sum R6. Then use geometry to find the exact value of
∫ 3
−3
f(r) dr.
y 1! 3! x 0.98!< <if( )=
y x 1! x 2.01< <if( )=
y 2x! 6+ 2 x 3< <if( )=
3!
1!
3
0
r
R6 =6∑i=1
f(ri) ∆r. Note that ∆r = 3−(−3)6
= 1. Therefore
R6 = f(−2) · 1 + f(−1) · 1 + f(0) · 1 + f(1) · 1 + f(2) · 1 + f(3) · 1
= (−1) · 1 + (−1) · 1 + 0 · 1 + 1 · 1 + 2 · 1 + 0 · 1 = 1.
Using geometry to find the exact value of
∫ 3
−3
f(r) dr gives
∫ 3
−3
f(r) dr = (2)(−1) + 12(1)(−1) + +1
2(3)(2) = 0.5.
31. Is the integral
∫ 1
1e
lnx dx positive, negative or zero? Explain (without integrating).
Note that lnx < 0 for all values of x in (1e, 1). Therefore
∫ 1
1e
lnx dx is negative.
32. Evaluate the integral
∫ 3
0
(1 − x3) dx using the definition of the definite integral as a
limit of a Riemann sum. (Do NOT use techniques from sections 5.3 or 5.4.)
Solution: Since a = 0 and b = 3, ∆x = 3−0n
= 3n
and then xi = 0 + i ·∆x = i · 3n
= 3in
.
So we get the Riemann sumn∑i=1
(1 − (3i
n)3)
3n. Simplifying gives 3
n
n∑i=1
(1 − 27i3
n3
)=
3n
n∑i=1
1− 3n
n∑i=1
27i3
n3 = 3n
n∑i=1
1− 81n4
n∑i=1
i3 = 3n·n− 81
n4 · n2(n+1)2
4= 3− 81(n+1)2
4n2 . Now take the
limit of this last expression as n→∞, getting 3− 814
= −694
. So
∫ 3
0
(1−x3) dx = −694
.
33. Evaluate the following definite integrals:
(a)
∫ 4
−1
|x− 1| dx∫ 4
−1
|x− 1| dx is the area under the two triangles in the following graph:
y x 1!=
0 y x 1! 1! x 4< <if( )< <
Their total area is (2 · 2 · 12) + (3 · 3 · 1
2) = 2 + 4.5 = 6.5.
(b)
∫ 2
0
(√4− x2 + 5
)dx∫ 2
0
(√
4− x2 + 5) dx =
∫ 2
0
√4− x2 dx+
∫ 2
0
5 dx.∫ 2
0
√4− x2 dx is the area of a quarter circle of radius 2, so
∫ 2
0
√4− x2 dx =
14· π(2)2 = π. Since
∫ 2
0
5 dx = 5(2− 0) = 10, we get that
∫ 2
0
(√
4− x2 + 5) dx =
π + 10.
34. Do problem 29 on page 374.
Note that f(x) =
∫ x
a
2t
tdt is an antiderivative of
2x
x. We want f(1) = 0, so let a = 1.
Therefore, f(x) =
∫ x
1
2t
tdt.
35. Findd
dx
(∫ x3
2x
e−t2
dt
). Hint: Look at problem #17 on page 373.
Rewrite
∫ x3
2x
e−t2
dt as
∫ 0
2x
e−t2
dt +
∫ x3
0
e−t2
dt = −∫ 2x
0
e−t2
dt +
∫ x3
0
e−t2
dt. Then
d
dx
(∫ x3
2x
e−t2
dt
)=
d
dx
(−∫ 2x
0
e−t2
dt+
∫ x3
0
e−t2
dt
)= −e−4x2 · (2) + e−x
6 · (3x2)
= 3x2e−x6 − 2e−4x2 .
36. Let F (x) =
∫ x
1
cos(πt2) dt. Find the equation of the line tangent to the graph of F (x)
at x = 1.
Solution: F (x) =
∫ x
1
cos(πt2) dt. F ′(x) = cos(πx2) by the 1st FTC. So F ′(1) =
cosπ = −1, so the slope of the line is −1. F (1) =
∫ 1
1
cos(πt2) dt = 0, so the point
of tangency is (1, 0). So the equation of the line is y = −(x − 1), which can also bewritten as y = −x+ 1.
37. Let g(x) =
∫ cosx
π
1
t4 + 1dt. Is g(x) is increasing or decreasing at x = 3π
2? Why?
Use the 1st FTC and the chain rule to get g′(x) =1
cos4 x+ 1· (− sinx). Then
g′(3π2
) =1
04 + 1(−(−1)) = 1. Since g′(3π
2) is positive, g(x) is increasing at x = 3π
2.
38. Let g(x) =
∫ x
0
(tet − 2et) dt for x ≥ 0. On what interval(s) is g(x) concave up?
g′(x) = xex − 2ex, so g′′(x) = xex + ex − 2ex = xex − ex. Set g′′(x) = 0, gettingex(x− 1) = 0⇒ x = 1. Then do a sign analysis of g′′(x):
ex + + + + + +
x− 1 −−− + + +
• →ex(x− 1) −−− + + +
1
Since g′′(x) > 0 on (1,+∞), g is concave up on (1,+∞).
39. Let f be a continuous function. Determine whether each of the following statements istrue or false. If a statement is true, briefly explain why. If a statement is false, explainwhy it’s false or else give an example that shows it’s false.
(a) If
∫ 2
0
f(x) dx = 6, then
∫ 4
0
f(x) dx = 12.
False. As an example, consider the function f(x) shown below. For this function,∫ 2
0
f(x) dx = 6, but
∫ 4
0
f(x) dx = 0.
y 6 3x!=
0 y 6 3x! 0 x 2< <if( )< <
6 3x! y 0 2 x 4< <if( )< <
(b) If
∫ 2
0
f(x) dx = 6 and g(x) = 2f(x), then
∫ 2
0
g(x) dx = 12.
True by the following property of integrals:
If c is a constant, then
∫ b
a
cf(x) dx = c
∫ b
a
f(x) dx.
So ∫ 2
0
g(x) dx =
∫ 2
0
2f(x) dx = 2
∫ 2
0
f(x) dx = 2 · 6 = 12.
(c) If a 6= b, then
∫ b
a
f(x) dx 6= 0.
False. As an example, consider the integral
∫ 4
0
f(x) dx, where f(x) is the function
shown in part (a). Note that a 6= b since 0 6= 4. However,
∫ 4
0
f(x) dx does equal
0.
(d)
∫ 2
0
f(x) dx ≤∫ 3
0
f(x) dx.
False. As an example, look again at the function f(x) shown in part (a). There∫ 2
0
f(x) dx = 6, but
∫ 3
0
f(x) dx = 6− 32
= 4.5. So
∫ 3
0
f(x) dx <
∫ 2
0
f(x) dx.
(e)
∫ b
a
f(x) dx =
∫ b
a
f(t) dt.
True. It doesn’t matter what we call the independent variable in a particularfunction f . For example, f(x) = x3 − 5 and f(t) = t3 − 5 represent the same
function. So
∫ b
a
f(x) dx =
∫ b
a
f(t) dt.
40. Evaluate the following:
(a) sin−1(−√
32
) = −π3
(b) tan(sin−1(− 1
6))
To find tan(sin−1(− 1
6)), start by letting α = sin−1
(− 1
6). Note that α is in the
fourth quadrant. We get the following triangle:
y 0 0 x 25< <if( )=
x 25 1! y 0< <if( )=
y1
25x! 0 x 25< <if
" #$ %& '
=
x 0=
y 0=
61
αx
Use the Pythagorean Theorem to get x =√
35. So tanα = − 1√35
, so
tan(sin−1(− 1
6))
= − 1√35
.
41. Simplify csc(tan−1 x). (Your answer should not contain any trig or inverse trig func-tions.)
Let θ = tan−1 x. First assume that x > 0. Then we can draw the following triangle:
y 1 1 x 3.75< <if( )=
x 3.75 1 y 4< <if( )=
y 0.75x 0.25+ 1 x 5< <if( )=
y 1.09x 0.09! 1 x 3.75< <if( )=
1
x
θ
The Pythagorean theorem tells us that the hypotenuse has length√
1 + x2. So
csc θ =1
sin θ=
√1 + x2
x.
Now assume that x < 0. Then θ has negative measure, so our triangle lies in the fourth
quadrant; once again the Pythagorean Theorem tells us that csc θ =
√1 + x2
x.
Finally, note that if x = 0 then csc(tan−1 0) = csc 0, which is undefined. So we canconclude the following:
csc(tan−1 x) =
√1 + x2
xfor all x 6= 0.
42. Suppose that f ′(x) = 5(1− x2)−12 and f(−1
2) = 15π
6. Find f(x).
Since f ′(x) = 5(1− x2)−12 = 5√
1−x2 we see that f(x) = 5 sin−1 x+ C.
Since f(−12) = 15π
6, we have:
5 sin−1
(−1
2
)+ C =
15π
6
⇒ 5(−π
6
)+ C =
15π
6
⇒ C =20π
6.
So f(x) = 5 sin−1 x+ 10π3.
43. Evaluate the following:
(a)9∑i=5
cos(πi)
9∑i=5
cos(πi) = cos(5π) + cos(6π) + cos(7π) + cos(8π) + cos(9π)
= (−1) + 1 + (−1) + 1 + (−1) = −1.
(b)n∑i=1
5
4n
Note that n is a constant in this sum. Son∑i=1
5
4n=
5
4n·
n∑i=1
1 =5
4n· n =
5
4.
44. Approximate the integral
∫ 3π2
0
cosx dx using R3. Then approximate the same integral
using L3.
Note that ∆x = π2. Then
R3 = f(π2) · π
2+ f(π) · π
2+ f(3π
2) · π
2
= cos(π
2
)· π
2+ cos(π) · π
2+ cos
(3π
2
)· π
2
= 0 · π2
+ (−1) · π2
+ 0 · π2
= −π2
andL3 = f(0) · π
2+ f(π
2) · π
2+ f(π) · π
2
= cos(0) · π2
+ cos(π
2
)· π
2+ cos(π) · π
2
= 1 · π2
+ 0 · π2
+ (−1) · π2
= 0.
45. Evaluate the integral
∫ 2
1
(x2 + 4x) dx using the definition of the definite integral as a
limit of a Riemann sum. (Do NOT use techniques from sections 5.3 or 5.4.)
Since a = 1 and b = 2, ∆x = 2−1n
= 1n. Moreover, xi = 1 + i ·∆x = 1 + i · 1
n= 1 + i
n.
So we get the Riemann sumn∑i=1
((1 + i
n
)2+ 4(1 + i
n))
1n
= 1n
n∑i=1
((1 + i
n
)2+ 4(1 + i
n))
= 1n
n∑i=1
(1 + 2i
n+ i2
n2 + 4 + 4in
)= 1n
n∑i=1
(5 + 6i
n+ i2
n2
)= 1
n
n∑i=1
5 + 1n
n∑i=1
6in
+ 1n
n∑i=1
i2
n2
= 5n
n∑i=1
1 + 6n2
n∑i=1
i+ 1n3
n∑i=1
i2 = 5n· n+ 6
n2 · n(n+1)2
+ 1n3 · n(n+1)(2n+1)
6
= 5 + 3 · n+1n
+ 2n3+3n2+n6n3 .
Now take the limit of this last expression as n → ∞, getting 5 + 3 + 13
= 253
. So∫ 2
1
(x2 + x) dx = 253
.
46. Express the following limit as a definite integral, using a = 2. Then express the samelimit as a different integral, this time using a = 0.
limn→∞
n∑i=1
√2 + (3i
n)5 · 3
n
To express the limit as a definite integral, we must put it in the form
∫ b
a
f(x) dx, so
we must find a, b and f(x).
(a) First let a = 2. It looks like ∆x = 3n. Therefore, since ∆x = b−a
nand a = 2, b
must equal 5. Since xi = a+ i∆x, we know that xi = 2 + 3in
. The function in the
integrand is√
2 + (3in
)5, so f(xi) =√
2 + (xi − 2)5. So f(x) =√
2 + (x− 2)5.
Hence the integral is
∫ 5
2
√2 + (x− 2)5 dx.
(b) Now let a = 0. Again, it looks like ∆x = 3n. Therefore, since ∆x = b−a
nand a = 0,
b must equal 3. xi = a+ i∆x, so here xi = 0 + 3in
= 3in
. Again the function in the
integrand is√
2 + (3in
)5, so f(xi) =√
2 + (xi)5. So f(x) =√
2 + x5. Hence the
integral is
∫ 3
0
√2 + x5 dx.
47. The graph of a function f(x) over the interval [−5, 5] is shown below. Suppose thatleft region has area A, the middle region has area B, and the right region has area C.Suppose we know that A = 5.6 and B = 1.3.
(a) What is
∫ 2
−5
f(x) dx?
∫ 2
−5
f(x)dx = −A+B = −5.6 + 1.3 = −4.3
(b) Suppose that we know that
∫ 5
−5
f(x) dx = −7.1. What is C?∫ 5
−5
f(x)dx = −7.1 ⇒ −A+B − C = −7.1 ⇒ −4.3− C = −7.1 ⇒ C = 2.8.
yx 5+( )( ) x 1+( ) x 2!( ) x 5!( )
80=
0 yx 5+( )( ) x 1+( ) x 2!( ) x 5!( )
801! x 2< <if" #
$ %< <
x 5+( )( ) x 1+( ) x 2!( ) x 5!( )
80y 0 5! x 1!< <if( )< <
x 5+( )( ) x 1+( ) x 2!( ) x 5!( )
80y 0 2 x 5< <if( )< <
48. Evaluate the following definite integrals:
(a)
∫ 3
−3
4√
9− x2 dx∫ 3
−3
4√
9− x2 dx = 4
∫ 3
−3
√9− x2 dx. The equation y =
√9− x2 represents the
semi-circle with radius 3, centered at the origin. So
∫ 3
−3
√9− x2 dx = 1
2· π(32) =
9π2
. So
∫ 3
−3
4√
9− x2 dx = 4 · 9π2
= 18π.
(b)
∫ 3
−2
(|x|+ 2
)dx∫ 3
−2
(|x|+2
)dx. There are several ways to compute this integral. One is to rewrite
the integral as
∫ 3
−2
|x| dx+
∫ 3
−2
2 dx. Use geometry to find
∫ 3
−2
|x| dx (see below),
getting 132
. For
∫ 3
−2
2 dx, we get 2(3−(−2)) = 10. So
∫ 3
−2
(|x|+2) dx = 132
+10 = 332
.
y x=
0 y x 2! x 3< <if( )< <
Another way to do the problem is to graph the function f(x) = |x|+2 (see below)and use geometry to find the area. The result is again 33
2.
y x 2+=
0 y x 2+ 2! x 3< <if( )< <
49. Let f(x) = sin(x3 + 1). Find a function F (x) such that F ′(x) = f(x) and F (0) = 7.
Note that F (x) =
∫ x
0
sin(t3 + 1) dt is an antiderivative of sin(x3 + 1). However,
F (0) = 0, not 7. So let F (x) = 7 +
∫ x
0
sin(t3 + 1) dt. This satisfies all the necessary
conditions.
50. Do problems 39 and 40 on page 426.
• Problem 39: F ′(x) =x2
1 + x3 .
• Problem 40: Use the 1st FTC and the chain rule: g′(x) =1− sin2 x
1 + sin4 x· cosx.