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from part (b), e1/(x+1) < eln(1+1/x) < e1/x , e1/(x+1) < 1 + 1/x < e1/x ,
ex/(x+1) < (1 + 1/x)x < e; by the Squeezing Theorem, lim (1 + 1/x)x = e.
x→+∞ (d) 45. Use the inequality ex/(x+1) < (1 + 1/x)x to get e < (1 + 1/x)x+1 so
(1 + 1/x)x < e < (1 + 1/x)x+1 . From Exercise 44(d) e − 1 + 1
50 50 < y (50), and from the graph y (50) < 0.06 0.2 0 100
0 46. F (x) = f (x), thus F (x) has a value at each x in I because f is continuous on I so F is continuous
on I because a function that is diﬀerentiable at a point is also continuous at that point CHAPTER 7 SUPPLEMENTARY EXERCISES
5. If the acceleration a = const, then v (t) = at + v0 , s(t) = 1 at2 + v0 t + s0
2 6. (a) Divide the base into n equal subintervals. Above each subinterval choose the lowest and highest
points on the curved top. Draw a rectangle above the subinterval going through the lowest point,
and another through the highest point. Add the rectangles that go through the lowest points to
obtain a lower estimate of the area; add the rectangles through the highest points to obtain an
upper estimate of the area. Supplementary Exercises 7 250 (b)
(c)
(a) 3
11
+=
24
4
5 −1 − (e)
8. n = 20: 24.4 cm, 23.1 cm (c) 7. n = 10: 25.0 cm, 22.4 cm not enough information (a)
(c) 3
4 =− (a)
(b)
(c) 10. 35
4 −1 1 dx + (f ) not enough information (b) not enough information
13
1
(d) 4(2) − 3 =
2
2 3 − π (3)2 /4 =
0 u = x2 , du = 2xdx; 1
2 −2 1 − x2 dx = 2(1) + π (1)2 /2 = 2 + π/2 −1 12
(x + 1)3/2
3 1
3
=−
2
2 (d) 5
1
+2=
2
2
not enough information
1 9. (b) −1 − 1 1
2 1
(103/2 − 1) − 9π/4
3
1 − u2 du = 0 1
π (1)2 /4 = π/8
2 y
1
0.8
0.6
0.4
0.2
x
0.2 0.6 1 11. The rectangle with vertices (0, 0), (π, 0), (π, 1) and (0, 1) has area π and is much too large; so is the
triangle with vertices (0, 0), (π, 0) and (π, 1) which has area π/2; 1 − π is negative; so the answer is
35π/128. 12. Divide ex + 3 into e2x to get
e2x
dx =
ex + 3 13. ex dx − 3 e2x
3ex
= ex − x
so
ex + 3
e +3
ex
dx = ex − 3 ln(ex + 3) + C
ex + 3 Since y = ex and y = ln x are inverse functions, their graphs are
symmetric with respect to the line y = x; consequently the areas A1
and A3 are equal (see ﬁgure). But A1 + A2 = e, so
e y
e
A1 1 x ln xdx +
1 e dx = A2 + A3 = A2 + A1 = e 1 0 A2
A3
1 14. (a) 1
n n k =1 1
lim
n→+∞ n k
=
n
n k =1 n f (x∗ )∆x where f (x) =
k
k =1 k
=
n 1 x1/2 dx =
0 2
3 √ e x x, x∗ = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
k 251 Chapter 7 (b) 1
n n k =1 1
lim
n→+∞ n
n (c)
k =1 n 15. k =1 1 x4 dx = =
0 1
5 n f (x∗ )∆x where f (x) = ex , x∗ = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
k
k
k =1
k/n e lim k =1
4 k
n k =1 n f (x∗ )∆x where f (x) = x4 , x∗ = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
k
k = ek/n
=
n n→+∞ n 4 k
n n
n→+∞ n 1 f (x∗ )∆x =
k = lim ex dx = e − 1. 0 k =1 1
is positive and increasing on the interval [1, 2], the left endpoint approximation
x
1
overestimates the integral of and the right endpoint approximation underestimates it.
x Since f (x) = (a) For n = 5 this becomes
1
1
1
1
1
+
+
+
+
<
0.2
1.2 1.4 1.6 1.8 2.0 2
1 1
1
1
1
1
1
dx < 0.2
+
+
+
+
x
1.0 1.2 1.4 1.6 1.8
2 (b) 1
dx = ln 2 is
x For general n the left endpoint approximation to
1 1
n n k =1 n k =1 n 1
=
1 + (k − 1)/n k =1
n 1
. This yields
n+k k =1 1
=
n+k−1
1
<
n+k 2
1 n−1
k =0 1
and the right endpoint approximation is
n+k
n−1 1
dx <
x k =0 1
which is the desired inequality.
n+k 1
1
1
1
=
so
≤ 0.1, n ≥ 5
(c) By telescoping, the diﬀerence is −
n 2n
2n
2n
(d) n ≥ 1, 000
16. The direction ﬁeld is clearly an even function, which means that the solution is even, its derivative
is odd. Since sin x is periodic and the direction ﬁeld is not, that eliminates all but x, the solution of
which is the family y = x2 /2 + C . 17. (a) n 1 · 2 + 2 · 3 + · · · + n(n + 1) = n k =1 =
n−1 (b)
k =1 k
9
−
n n2 17
lim
n→+∞ 2
3 n−1
n 2 (c) = j =1 n−1 1−
k =1 3 j=
j =1 1
n2 (a) n−1 k=
k =1 11
17
9
(n − 1) − 2 · (n − 1)(n) =
n
n2
2 i=1 1
2i + (2)(3) = 2
2 3 3 i+ n−1
;
n i=1 i=1 1
3 = 2 · (3)(4) + (3)(3) = 21
2 19 (k + 4)(k + 1)
k =0 k =1 1
1
1
n(n + 1)(2n + 1) + n(n + 1) = n(n + 1)(n + 2)
6
2
3 14 18. k k =1 17
=
2 2 i+
i=1 9
n n k2 + k (k + 1) = (k − 1)(k − 4) (b)
k =5 Supplementary Exercises 7 19. (a) 252 If u = sec x, du = sec x tan xdx, sec2 x tan xdx = if u = tan x, du = sec2 xdx,
(b)
20. 1
2 π /4 = 0 udu = u2 /2 + C2 = (tan2 x)/2 + C2 . x−1/3 1
2 tan2 x π /4
0 = 1
(1 − 0) = 1/2
2 x2/3 + 1dx; u = x2/3 + 1, du = 2 −1/3
x
dx
3 u1/2 du = u3/2 + C = (x2/3 + 1)3/2 + C
n n b 22. 1
(2 − 1) = 1/2 and
2 1 + x−2/3 dx =
3
2 udu = u2 /2 + C1 = (sec2 x)/2 + C1 ; They are equal only if sec2 x and tan2 x diﬀer by a constant, which is true. sec2 x 21. sec2 x tan xdx = b fk (x)dx = (a)
a k =1 fk (x)dx
k =1 a
b (b) yes; substitute ck fk (x) for fk (x) in part (a), and then use
a Theorem 7.5.4
x 23. (a)
1 fk (x)dx from
a 1
d...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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