Thomas Calculus 13th [Solutions]

Section 5.3 The Definite Integral 367
x
1 1
2 2 4 2 2n
cos cos n x cos cos n
n
Then U (sin x sin 2 x ... sin n x)
x x
x
2sin
2sin
2 4n
2n
cos cos cos cos
4n 2 4n 4n 2 4n
4nsin
4n
sin
4n
4n
(b) The area is
0
/2
cos cos 1 cos
4n
2 4n
2
sin x dx lim 1.
n
sin 1
4 n
4n
n
86. (a) The area of the shaded region is xi
m i which is equal to L.
i 1
n
(b) The area of the shaded region is xi
M i which is equal to U.
i 1
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part
of the figure and the first part of the figure. Thus this area is U L.
n
n
87. By Exercise 86, U L xi Mi xi m i where M i max { f ( x ) on the ith subinterval} and
i 1 i 1
n
n
m i min { f ( x ) on ith subinterval}. Thus U L ( Mi mi ) xi x i provided x i for each
i 1 i 1
n
n
i 1, , n. Since xi
xi
( b a ) the result, U L ( b a ) follows.
i 1 i 1
88. The car drove the first 150 miles in 5 hours and the second
150 miles in 3 hours, which means it drove 300 miles in
8 hours, for an average value of 300 mi/hr 37.5 mi/hr. In
8
terms of average value of functions, the function whose
30, 0 5
average value we seek is ( ) t
v t 50, 5 t 8 , and the
(30)(5) (50)(3)
average value is 37.5.
8
89-94. Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
f : x -> 1-x;
a : 0;
b : 1;
N : [4, 10, 20, 50];
P : [seq( RiemannSum( f(x), x a..b, partition n, method random, output plot ), n N )]:
display( P, insequence true);
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