Math 25-1,2
Exam 1
Fall 2011
1. (a) Precisely define the term antiderivative.
(b) What is true about any two antiderivatives of the same function?
(c) Why is the fact you state in part (b) of this question very important?
2. Suppose f is a continuous function such that the average value of f on the interval
[0, 2] is 4.
R2
(a) What is 0 f (x) dx?
(b) What conclusion can be drawn about the function f by applying the Mean Value
Theorem for Integrals?
3. Let f (x) = x4 .
(a) Estimate
R5
f (x) dx using Simpson’s Rule with four subintervals.
R5
(b) Suppose you were to estimate 3 f (x) dx using Simpson’s Rule. Use the error
bound formula given in class to find the minimum number of subintervals needed
to ensure an estimate within 10−10 of the actual value of the integral.
3
4. Evaluate one of the following two integrals:
Z 1
arcsin x dx
Z
x ln(3x) dx
1/3
0
5. Evaluate one of the following two integrals:
Z
cos3 x dx
1
Z
2x
dx
2x + 1
6. Evaluate one of the following two integrals:
Z
3x3 + 19x2 + 21x − 9
dx
x4 + 6x3 + 9x2
Z
2x2
dx
(x − 1)3
7. Given the improper integrals below, determine whether or not they converge. Explain
your answer.
Z 4
Z ∞ 2
2
2x + x2 sin x
(a)
dx
(b)
dx
4
4x6
2 (x − 2)
5
8. Let R be the√region under the graph of y = arctan x (and above the x−axis) from
x = 1 to x = 3.
(a) Suppose R is revolved around the x−axis to produce a solid. Write an integral
with respect to x that gives the volume of this solid.
(b) Suppose R is revolved around the line x = −2 to produce a solid. Write an
integral with respect to x that gives the volume of this solid.
(c) Suppose R is revolved around the line y = 3 to produce a solid. Write an integral
with respect to x that gives the volume of this solid.
Math 25
Exam 1 (Solutions)
Fall 2011
1. (a) An antiderivative of a function f (x) is another function F (x) satisfying F 0 (x) =
f (x).
(b) Any two antiderivatives of the same function differ by at most a constant.
(c) The fact in (b) tells us how to easily find all antiderivatives of a function, if
we know just a single antiderivative of that function (add an arbitrary constant
C). This idea is critical to the proof of the part of the Fundamental Theorem of
Calculus dealing with evaluation of integrals, so without it, we wouldn’t be able
to evaluate definite integrals using antiderivatives.
R2
1
2. (a) We are given 4 = 2−0
0 f (x) dx, so the integral must have value 8.
(b) By the Mean Value Theorem for Integrals, there must be at least one number
c ∈ [0, 2] such that f (c) = 4.
5−3
1
3. (a) Dividing [3, 5] into n = 4 subintervals, we see that b−a
n = 4 = 2 , so since
for all j, we have x0 = 3, x1 = 3.5, x2 = 4, x3 = 4.5, x4 = 5.
xj = a + j b−a
n
Now by the formula for Simpson’s Rule,
b−a
(f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 ))
3n
1 4
=
3 + 4(3.5)4 + 2(4)4 + 4(4.5)4 + 54 .
6
SIM P (4) =
(b) First, find M4 , the maximum value of |f 0000 (x)| on [3, 5]. We see that f 0 (x) =
4x3 , f 00 (x) = 12x2 , f 000 (x) = 24x nd f 0000 (x) = 24 which is at most 24 when
x ∈ [3, 5]. Therefore M4 = 24. Now
ES ≤
(b − a)5
(5 − 3)5
64
M
=
(24) =
.
4
4
4
180n
180 · n
15n4
Since we want ES ≤ 10−10 , we can set
64
≤ 10−10
15n4
q
64
4
10
10
and solve for n to get n ≥ 15 (10 ), i.e. n ≥ 4 64
15 (10 ). (Of course n needs to
be an even number as well.)
4. For the first integral, use integration by parts with u = arcsin x and dv = dx. Then
1
du = √1−x
dx and v = x so by the integration by parts formula,
2
Z
1
Z
arcsin x dx =
0
Z
u dv = uv −
v du = x arcsin x|10 −
Z
0
1
√
x
dx
1 − x2
To handle the integral at the right, perform a u−substitution with u = 1 − x2 , du =
2
−2xdx so −1
2 du = xdx (and change the limits from x = 0 to x = 1 to u = 1 − 0 = 1
Math 25
Exam 1 (Solutions)
to u = 1 − 12 = 0) to obtain
Z 1
Z
arcsin x dx = x arcsin x|10 −
Fall 2011
0
−1 1
√ du
2 u
1
√
= (1 arcsin 1 − 0 arcsin 0) − (− u)|01
π
= − (0 + 1)
2
π
= − 1.
2
0
For the second integral, use integration by parts with u = ln 3x and dv = x dx. Then
2
1
du = 3x
3 dx = x1 dx and v = x2 so by the integration by parts formula,
Z
1
Z
x ln(3x) dx =
1/3
Z
u dv = uv −
1
Z 1 2
x 1
x2
ln(3x) −
dx
v du =
2
1/3 2 x
1/3
Z 1
ln 3 2
x
=
− ·0 −
dx
2
9
1/3 2
1
ln 3
x2 1
1
ln 3
=
−
−
−
=
2
4 1/3
2
4 36
=
ln 3 2
− .
2
9
5. Rewrite the first integral as
Z
Z
Z
cos3 x dx = cos2 x cos x dx = (1 − sin2 x) cos x dx
and perform the u−substitution u = sin x, du = cos x dx to obtain
Z
u3
sin3 x
(1 − u2 ) du = u −
+ C = sin x −
+ C.
3
3
For the second integral, use the substitution u = 2x + 1, du = 2dx and x = u−1
2 to
obtain
Z
Z Z
2x
u−11
1
1
1
1
dx =
du =
−
du = u − ln |u| + C
2x + 1
2 u
2 2u
2
2
1
1
= (2x + 1) − ln |2x + 1| + C.
2
2
6. Both these integrals use partial fractions. For the first integral, start by factoring the
denominator:
x4 + 6x3 + 9x2 = x2 (x2 + 6x + 9) = x2 (x + 3)2
Now the “guessed” form of the decomposition is
3x3 + 19x2 + 21x − 9
B
D
A
C
= + 2+
+
;
4
3
2
x + 6x + 9x
x
x
x + 3 (x + 3)2
Math 25
Exam 1 (Solutions)
Fall 2011
add the fractions on the right-hand side and clear denominators to obtain
3x3 + 19x2 + 21x − 9 = Ax(x + 3)2 + B(x + 3)2 + Cx2 (x + 3) + Dx2 .
Plug in x = 0 to both sides to obtain −9 = 9B, i.e. B = −1. Plug in x = −3 to both
sides to obtain −81 + 171 − 63 − 9 = 9D, i.e. 18 = 9D, i.e. D = 2. Now our equation
is
3x3 + 19x2 + 21x − 9 = Ax(x + 3)2 − (x + 3)2 + Cx2 (x + 3) + 2x2 ;
plugging in x = 1 to both sides yields 3 + 19 + 21 − 9 = 16A − 16 + 4C + 2, i.e.
48 = 16A + 4C, i.e. 12 = 4A + C. Plugging in x = −1 to both sides yields −3 + 19 −
21 − 9 = −4A − 4 + 2C + 2, i.e. −12 = −4A + 2C. Now solve the two equations
12 = 4A + C,
−12 = −4A + 2C
for A and C to get A = 3, C = 0. Thus the partial fraction decomposition is
3x3 + 19x2 + 21x − 9
3
1
2
= − 2+
4
3
2
x + 6x + 9x
x x
(x + 3)2
so the integral is
Z
Z 3x3 + 19x2 + 21x − 9
3
1
2
dx =
−
+
dx
x4 + 6x3 + 9x2
x x2 (x + 3)2
1
2
= 3 ln |x| + −
+ C.
x x+3
For the second integral, the denominator is already factored. The “guessed” form of
the decomposition is
A
B
C
2x2
=
+
+
;
3
2
(x − 1)
x − 1 (x − 1)
(x − 1)3
add the fractions on the right-hand side and clear denominators to obtain
2x2 = A(x − 1)2 + B(x − 1) + C;
substitute in x = 1 to both sides to get 2 = C. Now the equation reduces to
2x2 = A(x − 1)2 + B(x − 1) + 2;
plug in x = 0 to get 0 = A − B + 2, i.e. A = B − 2, and plug in x = 2 to get
8 = A + B + 2. Therefore A = 2 and B = 4 so
2x2
2
4
2
=
+
+
(x − 1)3
x − 1 (x − 1)2 (x − 1)3
and the integral is therefore
Z
Z 2x2
2
4
2
dx =
+
+
dx
(x − 1)3
x − 1 (x − 1)2 (x − 1)3
4
1
= 2 ln |x − 1| −
−
+ C.
x − 1 (x − 1)2
Math 25
Exam 1 (Solutions)
Fall 2011
7. (a) Notice that the integrand has a vertical asymptote at x = 2, the left-hand limit
of this improper integral. Rewriting this as a limit of a definite integral and
evaluating, we see that
Z 4
Z 4
2
2
dx
=
lim
dx
4
+
(x
−
2)
(x
−
2)4
b→2
2
b
−2 4
= lim
b→2+ 3(x − 2)3 b
−2
−2
= lim
−
3(b − 2)3
b→2+ 3(4 − 2)3
= ∞.
Therefore the integral diverges.
(b) Notice that 2x2 + x2 sin x = x2 (2 + sin x) ≤ x2 (2 + 1) = 3x2 . Dividing through
by 4x6 , we see that
3x2
3
2x2 + x2 sin x
≤
= 4.
6
6
4x
4x
4x
We know that
Z ∞
Z ∞
1
3
3
dx =
dx
4
4x
4 5 x4
5
R∞
converges (it is a constant times the convergent integral 5 x1p dx where p = 4 >
1), so by the Comparison Test for Integrals,
Z ∞ 2
2x + x2 sin x
dx converges.
4x6
5
8. (a) Since the direction of integration is parallel to the axis of revolution, we use
washers (in this case, disks as the solid has no “hole”): A(x) = πr2 = π arctan2 x
so
Z √
3
π arctan2 x dx.
V =
1
Note: the notation arctan2 x means (arctan x)2 .
(b) Now the direction of integration is perpendicular to the axis of revolution, so we
use the shell method. A(x) = 2πrh where r = x−(−2) = x+2 and h = arctan x;
we have
Z √
3
2π(x + 2) arctan x dx.
V =
1
(c) As in part (a), the direction of integration is parallel to the √
axis of revolution,
so we use washers as our cross-sections. Notice that arctan 3 = π/3 < 3, so
the axis of revolution is above the region R. Therefore, we have R = 3 and
r = 3 − arctan x, so A(x) = πR2 − πr2 = 9π − π(3 − arctan x)2 . Therefore
Z √3
V =
9π − π(3 − arctan x)2 dx.
1