CHALLENGE PROBLEMS
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1
CHALLENGE PROBLEMS:
CHAPTER 5
A Click here for answers.
1. If x sin ␲ x 苷
y
x2
0
S
Click here for solutions.
f 共t兲 dt, where f is a continuous function, find f 共4兲.
2. In this problem we approximate the sine function on the interval 关0, ␲兴 by three quadratic func-
;
tions, each of which has the same zeros as the sine function on this interval.
(a) Find a quadratic function f such that f 共0兲 苷 f 共␲兲 苷 0 and which has the same maximum
value as sin on 关0, ␲兴.
(b) Find a quadratic function t such that t共0兲 苷 t共␲兲 苷 0 and which has the same rate of change
as the sine function at 0 and ␲.
(c) Find a quadratic function h such that h共0兲 苷 h共␲兲 苷 0 and the area under h from 0 to ␲ is the
same as for the sine function.
(d) Illustrate by graphing f, t, h, and the sine function in the same viewing rectangle 关0, ␲兴 by
关0, 1兴. Identify which graph belongs to each function.
3. Show that
1
1
7
2
.
艋y
dx 艋
4
1 1 ⫹ x
17
24
4. Suppose the curve y 苷 f 共x兲 passes through the origin and the point 共1, 1兲. Find the value of the
integral x01 f ⬘共x兲 dx.
5. Find a function f such that f 共1兲 苷 ⫺1, f 共4兲 苷 7, and f ⬘共x兲 ⬎ 3 for all x, or prove that such a
function cannot exist.
2
3
; 6. (a) Graph several members of the family of functions f 共x兲 苷 共2cx ⫺ x 兲兾c for c ⬎ 0 and look at
the regions enclosed by these curves and the x-axis. Make a conjecture about how the areas of
these regions are related.
(b) Prove your conjecture in part (a).
(c) Take another look at the graphs in part (a) and use them to sketch the curve traced out by the
vertices (highest points) of the family of functions. Can you guess what kind of curve this is?
(d) Find the equation of the curve you sketched in part (c).
t共x兲
1
cos x
7. If f 共x兲 苷 y
dt, where t共x兲 苷 y 关1 ⫹ sin共t 2 兲兴 dt, find f ⬘共␲兾2兲.
0
0
s1 ⫹ t 3
8. If f 共x兲 苷
x0x x 2 sin共t 2 兲 dt, find
f ⬘共x兲.
9. Find the interval 关a, b兴 for which the value of the integral xab 共2 ⫹ x ⫺ x 2 兲 dx is a maximum.
10000
兺
10. Use an integral to estimate the sum
si.
i苷1
11. (a) Evaluate x0n 冀x冁 dx, where n is a positive integer.
(b) Evaluate xab 冀x冁 dx, where a and b are real numbers with 0 艋 a ⬍ b.
12. Find
d2
dx 2
x
冉
y y
0
sin t
1
冊
s1 ⫹ u 4 du dt.
x
13. If f is a differentiable function such that y f 共t兲 dt 苷 关 f 共x兲兴 2 for all x, find f .
0
2
14. A circular disk of radius r is used in an evaporator and is rotated in a vertical plane. If it is to be
partially submerged in the liquid so as to maximize the exposed wetted area of the disk, show that
the center of the disk should be positioned at a height r兾s1 ⫹ ␲ 2 above the surface of the liquid.
Thomson Brooks-Cole copyright 2007
2
2
x
15. Prove that if f is continuous, then y f 共u兲共x ⫺ u兲 du 苷
0
x
冉
y y
0
u
0
冊
f 共t兲 dt du.
16. The figure shows a region consisting of all points inside a square that are closer to the center than
to the sides of the square. Find the area of the region.
2
FIGURE FOR PROBLEM 16
17. Evaluate lim
nl⬁
冉
冊
1
1
1
.
⫹
⫹ ⭈⭈⭈ ⫹
sn sn ⫹ 1
sn sn ⫹ 2
sn sn ⫹ n
2
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CHALLENGE PROBLEMS
ANSWERS
S
Solutions
1. 兾2
5. Does not exist
11. (a) 共n 1兲n兾2
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9. 关1, 2兴
(b) 冀b冁共2b 冀b冁 1兲 12 冀a冁共2a 冀a冁 1兲
13. f 共x兲 苷 2 x or f 共x兲 苷 0
1
7. 1
1
2
17. 2(s2 1)
CHALLENGE PROBLEMS
SOLUTIONS
E
Exercises
x2
0
1. Differentiating both sides of the equation x sin πx =
f (t) dt (using FTC1 and the Chain Rule for the right side)
gives sin πx + πx cos πx = 2xf x2 . Letting x = 2 so that f x2 = f (4), we obtain
sin 2π + 2π cos 2π = 4f (4), so f (4) = 14 (0 + 2π · 1) =
π
2.
3. For 1 ≤ x ≤ 2, we have x4 ≤ 24 = 16, so 1 + x4 ≤ 17 and
2
1
2
1
1
dx ≥
1 + x4
2
1
dx <
1 + x4
2
2
1
≤
17
1
1
1
≥
. Thus,
1 + x4
17
1
1
1
1
dx =
. Also 1 + x4 > x4 for 1 ≤ x ≤ 2, so
< 4 and
17
17
1 + x4
x
1
x−4 dx =
1
2
x−3
−3
1
=−
1
7
1
+ =
. Thus, we have the estimate
24
3
24
1
7
.
dx ≤
1 + x4
24
5. Such a function cannot exist. f 0 (x) > 3 for all x means that f is differentiable (and hence continuous) for all x. So
by FTC2,
4
1
4
1
f 0 (x) dx = f (4) − f (1) = 7 − (−1) = 8. However, if f 0 (x) > 3 for all x, then
f 0 (x) dx ≥ 3 · (4 − 1) = 9 by Comparison Property 8 in Section 5.2.
Another solution: By the Mean Value Theorem, there exists a number c ∈ (1, 4) such that
f 0 (c) =
7 − (−1)
8
f (4) − f (1)
=
=
4−1
3
3
⇒ 8 = 3f 0 (c). But f 0 (x) > 3 ⇒ 3f 0 (c) > 9, so such a function
cannot exist.
7. f (x) =
g(x)
0
1
√
dt, where g(x) =
1 + t3
1
f 0 (x) =
1 + [g(x)]
g
π
2
=
0
0
1
g 0 (x) =
3
1 + sin t2
cos x
0
dt. Using FTC1 and the Chain Rule (twice) we have
1 + sin cos2 x
1 + [g(x)]
dt = 0, so f 0
1 + sin t2
π
2
(− sin x). Now
3
=
√ 1
(1
1+0
+ sin 0)(−1) = 1 · 1 · (−1) = −1.
9. f (x) = 2 + x − x2 = (−x + 2)(x + 1) = 0 ⇔ x = 2 or x = −1. f (x) ≥ 0 for x ∈ [−1, 2] and f (x) < 0
b
a
everywhere else. The integral
2 + x − x2 dx has a maximum on the interval where the integrand is positive,
which is [−1, 2]. So a = −1, b = 2. (Any larger interval gives a smaller integral since f (x) < 0 outside [−1, 2].
Any smaller interval also gives a smaller integral since f (x) ≥ 0 in [−1, 2].)
11. (a) We can split the integral
n
0
[[x]] dx into the sum
n
i=1
i
i−1
[[x]] dx . But on each of the intervals [i − 1, i) of
integration, [[x]] is a constant function, namely i − 1. So the ith integral in the sum is equal to
(i − 1)[i − (i − 1)] = (i − 1). So the original integral is equal to
Thomson Brooks-Cole copyright 2007
(b) We can write
Now
b
0
b
a
[[x]] dx =
[[x]] dx =
[[b]]
0
b
0
[[x]] dx −
[[x]] dx +
b
[[b]]
a
0
n−1
n
(i − 1) =
i=1
i=
i=1
(n − 1)n
.
2
[[x]] dx.
[[x]] dx. The first of these integrals is equal to 12 ([[b]] − 1) [[b]], by part (a),
and since [[x]] = [[b]] on [[[b]] , b], the second integral is just [[b]] (b − [[b]]). So
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CHALLENGE PROBLEMS
b
0
[[x]] dx = 12 ([[b]] − 1) [[b]] + [[b]] (b − [[b]]) =
a
0
[[x]] dx =
1
2
1
2
[[b]] (2b − [[b]] − 1) and similarly
b
a
[[x]] dx =
[[a]] (2a − [[a]] − 1). Therefore,
13. Differentiating the equation
x
0
x
0
[[b]] (2b − [[b]] − 1) −
1
2
[[a]] (2a − [[a]] − 1).
f (t) dt = [f(x)]2 using FTC1 gives f (x) = 2f(x)f 0 (x) ⇒
f (x)[2f 0 (x) − 1] = 0, so f (x) = 0 or f 0 (x) = 12 . f 0 (x) =
the original equation to get
1
2
1
t
2
1
x
2
+ C dt =
+C
2
⇒ f (x) = 12 x + C. To find C we substitute into
1
2
⇔
1 2
x
4
+ Cx = 14 x2 + Cx + C 2 . It follows that
C = 0, so f (x) = 12 x. Therefore, f (x) = 0 or f (x) = 12 x.
15. Note that
d
dx
x
u
x
f (t) dt du
0
f (t) dt by FTC1, while
=
0
0
⎤
⎡x
⎡
d
d ⎣
⎣x
f (u)(x − u) du⎦ =
dx
dx
0
x
0
⎤
⎤
⎡x
d
⎣ f (u)u du⎦
f (u) du⎦ −
dx
0
x
=
0
Hence,
17. lim
n→∞
x
0
f (u)(x − u) du =
u
0
= lim
1
n
n
+
n+1
= lim
1
n
1
+
1 + 1/n
n→∞
1
n→∞ n
n
f
= lim
1
=
0
i=1
i
n
f (u) du
0
f (t) dt du + C. Setting x = 0 gives C = 0.
1
1
1
+√ √
+··· + √ √
√ √
n n+n
n n+1
n n+2
n→∞
Thomson Brooks-Cole copyright 2007
x
0
x
f(u) du + xf(x) − f(x)x =
n
+··· +
n+2
n
n+n
1
1
+ ··· + √
1+1
1 + 2/n
1
where f (x) = √
1+x
√
1
√
dx = 2 1 + x
1+x
1
0
=2
√
2−1