Math 1105: Calculus I (Math/Sci majors)
MWF 11am / 12pm, Campion 235
Written homework 2
§5.3: p. 376(107,112), §5.4: p. 383(57,62,66), §5.5: p. 393(98,103,107)
p. 376, 107. What value of b > −1 maximizes the integral
Z
b
x2 (3 − x)dx?
−1
Solution. Using the Fundamental Theorem of Calculus (part 2), we compute:
Z
b
Z
2
b
x (3 − x)dx =
−1
2
3
3x − x dx =
−1
b
3x3 x4 −
3
4 −1
b
x4 b4
1
b4 5
3
3
3
= x −
= b − − (−1 − ) = b − + .
4 −1
4
4
4
4
This integral is maximized exactly for the value of b that maximizes b3 − b4 /4 + 5/4, which we compute
using calculus (by finding values of b for which the derivative vanishes). We have:
0
b4 5
3
b − +
= 3b2 − b3 = 0,
4
4
exactly when 3b2 = b3 , i.e. when b = 3. As a function of b, b3 − b4 /4 + 5/4 has a single critical point at
b = 3, and since the coefficient of the highest degree term, b4 , is negative, it approaches negative infinity
as b goes to plus or minus infinity. We conclude that there is a unique maximum at b = 3.
p. 376, 112. Show that the sine integral S(x) =
2S 00 (x) + xS 000 (x) = 0.
Rx
0
sin t
dt
t
satisfies the differential equation xS 0 (x) +
Solution. By the Fundamental Theorem of Calculus (part 1), we have
S 0 (x) =
sin x
.
x
This allows us to compute higher derivatives of S(x):
0
x cos x − sin x
x2
0
x cos x − sin x
x2 (x cos x − sin x)0 − (x cos x − sin x)2x
000
S (x) =
=
x2
x4
x2 (cos x − x sin x − cos x) − (x cos x − sin x)2x
=
x4
−x3 sin x − 2x2 cos x + 2x sin x
=
x4
00
S (x) =
sin x
x
=
1
We compute the relevant differential equation:
xS 0 (x)+2S 00 (x) + xS 000 (x) =
3
sin x
x cos x − sin x
−x sin x − 2x2 cos x + 2x sin x
x
+2
+x
x
x2
x4
cos x
sin x
sin x
cos x
= sin x + 2
− 2 2 − sin x − 2
+ 2 2 = 0.
x
x
x
x
p. 383, 57. Consider the function f (x) = ax(1 − x) on the interval [0, 1], where a is a positive real
number.
(a) Find the average value of f as a function of a.
(b) Find the points at which the value of f equals its average value and prove that the are independent
of a.
Solution (a). We compute:
Z 1
Z 1
1
¯
ax(1 − x)dx =
ax − ax2 dx
f=
1−0 0
0
2
3 1
x
x = a −a
2
3 0
a
a a
= − = .
2 3
6
Thus the average value of f on [0, 1] is a/6.
Solution (b). We solve the equation f (c) = f¯ for c:
a
ac(1 − c) = .
6
After dividing by a and distributing, we see:
1
c − c2 = .
6
Evidently, either of the two solutions to this quadratic equation are independent of a, but we solve them
in any case: Rearranging the above equation, we find
1
0 = c2 − c + ,
6
and the general solution to the quadratic equation provides:
q
1 ± 1 − 4 16
1 ± √13
c=
=
.
2
2
p. 383, 62.
(a) Consider the curve y = 1/x for x ≥ 1. For what value of b > 0 does the region bounded by this
curve and the x-axis on the interval [1, b] have an area of 1?
(b) Consider the curve y = 1/xp where x ≥ 1 and p < 2 with p 6= 1. For what value of b (as a function
of p) does the region bounded by this curve and the x-axis on the interval [1, b] have unit area?
(c) Is b(p) in part (b) an increasing or decreasing function of p? Explain.
Solution (a). Using the Fundamental Theorem of Calculus (part 2) we compute
Z b
dx
= log x|b1 = log b − log 1 = log b,
1 x
which is equal to 1 when b = e.
Solution (b). Using the Fundamental Theorem of Calculus (part 2) we compute
b
Z b
dx
1
x1−p b1−p
1=
−
.
=
=
p
1−p 1 1−p 1−p
1 x
Rearranging, we find:
= (1 − p) 1 +
1
b
= 1 − p + 1 = 2 − p.
1−p
Taking both sides to the exponent 1/(1 − p) (which is legit because p 6= 1 and p < 2), we have
1−p
b = (2 − p)1/(1−p) .
Solution (c). In order to determine if b(p) is increasing or decreasing, we must compute the derivative.
To make things simpler, we will compute the derivative of log b(p). Since b(p) > 0, the function b(p) is
increasing exactly when log b(p) is increasing.
0
1
log(2 − p)
1−p
1
1
= −(1 − p)−2 (−1) log(2 − p) +
· (−1)
1−p
2−p
(2 − p) log(2 − p) + (p − 1)
=
(1 − p)2 (2 − p)
b0 (p)
= (log b(p))0 =
b(p)
It remains to determine the sign of (2 − p) log(2 − p) + (p − 1) for p < 2 (and p 6= 1). To make this easier
to look at, let x = 2 − p, and note that p < 2 and p 6= 1 implies that x > 0 and x 6= 1. Now we have
(2 − p) log(2 − p) + (p − 1) = x log x + (1 − x).
To make our analysis easier, let g(x) = x log x + 1 − x, the last expression. Note that (2 − p) log(2 −
p) + (p − 1) > 0 exactly when g(x) > 0 above.
In fact, we can use calculus to show that g(x) ≥ 0 for x > 0, with g(x) = 0 if and only if x = 1: Using
product rule, we can see
1
g 0 (x) = log x + x · − 1 = log x,
x
which is negative for x < 1 and positive for x > 1. This implies that g is strictly decreasing for x < 1
and strictly increasing for x > 1, which means that g(x) > g(1) = 0 for all x.
p. 383, 66. Suppose that f 0 is a continuous function for all real numbers. Show that the average
(a)
value of the derivative on an interval [a, b] is f 0 = f (b)−f
. Interpret the result in terms of secant lines.
b−a
Solution. We will compute the average value of the derivative on [a, b], using the Fundamental
Theorem of Calculus (part 2) to compute the definite integral: Since f is an anti-derivative of f 0 , we
have:
Z b
f (b) − f (a)
1
1
0
f (x)|ba =
,
f =
f 0 (x)dx =
b−a a
b−a
b−a
as desired. The quantity on the right is evidently the slope of the line between the points (a, f (a)) and
(b, f (b)) on the graph of f , so we conclude that the average value of the derivative on an interval is given
by the slope between the points on the graph between the endpoints of the interval.
Z
p. 393, 98. Evaluate 2(f 2 (x) + 2f (x))f (x)f 0 (x)dx. (Note that f p (x) is the pth power of f (x).)
Solution. We have
Z
2
0
2(f (x) + 2f (x))f (x)f (x)dx =
Z
2(f 3 (x) + 2f 2 (x))f 0 (x)dx.
We perform a substitution using u = f (x), so that du = f 0 (x)dx. The above integrals become
Z
Z
3
2
2(u + 2u )du = 2 u3 + 2u2 du
4
u
u3
=2
+2
+C
4
3
u4 4u3
+
+C
=
2
3
f 4 (x) 4f 3 (x)
+
+ C.
=
2
3
R
p. 393, 103. Consider the integral I = sin2 x cos2 xdx.
(a) Find I using the identity sin 2x = 2 sin x cos x.
(b) Find I using the identity cos2 x = 1 − sin2 x.
(c) Confirm that the results in parts (a) and (b) are consistent and compare the work involved in each
method.
Z
Solution (a). Since sin 2x = 2 sin x cos x, we have I =
sin2 2x
1
dx =
4
4
Z
substitution u = 2x (and du = 2dx), the above becomes
Z
1
I=
sin2 udu,
8
which we solve using the trig identity sin2 u = (1 − cos 2u)/2:
Z
Z
1
1
1 − cos 2u
2
I=
sin u du =
du
8
8
2
Z
1
1
sin 2u
=
1 − cos 2u du =
u−
+C
16
16
2
x sin 4x
= −
+ C.
8
32
Solution (b). Since cos2 x = 1 − sin2 x, we have
Z
Z
Z
2
2
2
I = sin x(1 − sin x)dx = sin x dx − sin4 x dx.
sin2 2x dx. Using the
Each of these we compute using the identity sin2 x = (1 − cos 2x)/2:
Z
Z
Z
1 − cos 2x
1
2
sin x dx =
dx =
1 − cos 2x dx
2
2
sin 2x
x sin 2x
1
x−
+C = −
+ C, and
=
2
2
2
4
2
Z
Z Z
1 − cos 2x
1
4
sin x dx =
dx =
1 − 2 cos 2x + cos2 2x dx
2
4
Z
1
1 + cos 4x
=
dx
1 − 2 cos 2x +
4
2
Z
1
3
cos 4x
=
− 2 cos 2x +
dx
4
2
2
sin 4x
1 3
x − sin 2x +
+C
=
4 2
8
3x sin 2x sin 4x
=
−
+
+ C.
8
4
32
Putting these together, we obtain:
x sin 2x
3x sin 2x sin 4x
x sin 4x
−
−
+
+ C.
I=
−
+C = −
2
4
8
4
32
8
32
Solution (c). Parts (a) and (b) give the same answer, but part (b) is a lot more bookkeeping!
Z
dx
p
p. 393, 107. Evaluate
.
√
1+ 1+x
√
√
1
Solution. We perform the substitution u = 1 + x, so that du = 12 √1+x
dx, and dx = 2 1 + x du:
Z √
Z
Z
2 1+x
dx
u
p
√
=
du = 2 √
du.
√
1+u
1+u
1+ 1+x
The last we evaluate by performing the substitution v = u + 1, so that du = dv and we have
3/2
Z
Z
Z
√
√
u
v−1
1
2v
4√
√ dv = 2
2 √
du = 2
v − √ dv = 2
v(v − 3) + C.
−2 v +C =
3
3
v
v
1+u
Substituting back in u and then x, we have:
Z
q
√
√
dx
4
4 √
p
= (u − 2) 1 + u + C =
1+x−2
1 + 1 + x + C.
√
3
3
1+ 1+x
Problem A. Recall the function
f (x) =

 1

0
if x is rational
if x is irrational.
In Problem 83 from §5.2 (on HW1) you showed that some Riemann sums, with width approaching 0,
are equal to 1 on [0, 1]. Show that a different choice of x∗k in these Riemann sums will yield the value 0.
Conclude that this function f is not integrable. (HINT: Between each pair of numbers there exists an
irrational number.)
Solution A. For any partition 0 = x0 < x1 < . . . < xn = 1, in order to form a Riemann sum we may
choose the points x∗k from the interval [xk−1 , xk ] freely, for k = 1, . . . , n. By the hint, for each k there is
an irrational in the interval [xk−1 , xk ], which we denote αk , and we make the choice x∗k = αk . Forming
the Riemann sum, we have
n
n
n
X
X
X
∗
f (xk )∆xk =
f (αk )∆xk =
0 · ∆k = 0,
k=1
k=1
k=1
where f (αk ) = 0 for each k = 1, . . . , n since αk is irrational. Thus, with our choice of x∗k ’s, the Riemann
sum is 0 for each n. On HW1 we found a sequence of Riemann sums, with width approaching 0, that
were always equal to 1. This implies that the limit
n
X
lim
f (x∗k )∆xk
∆→0
k=1
is not independent of the choices made to form the Riemann sums, and we conclude that f is not
integrable.