Thomas Calculus 13th [Solutions]

512 Chapter 7 Integrals and Transcendental Functions
(c)
The approximation y = x for ln(1 + x) is best for smaller
positive values of x; in particular for 0 x 0.1 in the
graph. As x increases, so does the error x ln(1 + x). From
the graph an upper bound for the error is 0.5 ln(1 + 0.5)
0.095; i.e., E( x ) 0.095 for 0 x 0.5. Note from the
graph that 0.1 ln(1 + 0.1) 0.00469 estimates the error in
replacing ln(1 + x) by x over 0 x 0.1. This is consistent
with the estimate given in part (b) above.
x
x
58. (a) f ( x) e f ( x) e ; L( x) f (0) f (0)( x 0) L( x) 1 x
(b) f(0) = 1 and L(0) = 1 error = 0;
x
(c) Since y e 0, the tangent line approximation
x
always lies below the curve y e . Thus L(x) = x + 1
x
never overestimates e .
0.2
f (0.2) e 1.22140 and L(0.2) = 1.2 error 0.02140
59. Note that y = ln x and
ln a
e y
dy
0
rectangle
y
e
x are the same curve;
a
1 ln x dx are under the curve between 1 and a;
area to the left of the curve between 0 and ln a. The sum of these areas is equal to the area of the
a ln a
ln x dx e y
dy a ln a.
1 0
x
x
60. (a) y e y e 0 for all x the graph of y
(b) area of the trapezoid ABCD <
(c)
x
e is always concave upward
ln b
e x
dx area of the trapezoid AEFD 1 ( AB CD)(ln b ln a)
ln a 2
ln b x e
ln a e
ln b
(ln ln ).
ln a e dx 2
b a Now 1 2 ( AB CD ) is the height of the midpoint (ln a ln b)/2
M e
since the curve containing the points B and C is linear
(ln a ln b)/2
(ln ln ) ln b x
ln a ln b
e b a e dx e e
ln a
2
(ln b ln a )
ln b ln b
ln ln
e x
dx e x
e b
e a
b a , so part (b) implies that
ln a
ln a
ln a
ln b
(ln a ln b)/2 (ln a ln b)/2
e (ln b ln a) b a e e (ln b ln a)
e
b a a b
2 ln b ln a 2
ln a/2 ln b/2 ln a ln b
e e b a a b e e b a a b ab b a a b
ln b ln a 2 ln b ln a 2 ln b ln a 2
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