Thomas Calculus 13th [Solutions]

366 Chapter 5 Integration
(c) av( f )
1
b
( )
1
b
( )
b a a
f x dx b a a
g x dx since f ( x ) g ( x ) on [ a, b ], and
1
b
( ) av( ).
b a a g x dx g
Therefore, av( f ) av( g).
83. (a) U max1 x max2
x max n x where max 1 f ( x1 ), max2
f ( x2
) , , max n f ( x n ) since f is
increasing on [ a, b]; L min1 x min2
x min n x where min 1 f ( x0 ), min2
f ( x1
), ,
min n f ( x n 1)
since f is increasing on [ a, b ]. Therefore
U L (max1 min 1) x (max2 min 2) x (maxn
min n)
x
( f ( x1 ) f ( x0 )) x ( f ( x2 ) f ( x1 )) x ( f ( xn ) f ( xn 1)) x ( f ( xn
) f ( x0
)) x
( f ( b) f ( a)) x.
(b) U max1 x1 max2 x2
max n x n where max 1 f ( x1 ), max 2 f ( x2
) , , max n f ( x n ) since f
is increasing on [ a, b]; L min1 x1 min 2 x2
... min n x n where min 1 f ( x0 ), min 2 f ( x1
), ,
min n f ( x n 1)
since f is increasing on [ a, b ]. Therefore
U L (max1 min 1) x1 (max2 min 2 ) x2
(maxn min n)
xn
( f ( x1 ) f ( x0 )) x1 ( f ( x2 ) f ( x1 )) x2 ( f ( xn ) f ( xn 1))
xn
( f ( x1 ) f ( x0 )) xmax ( f ( x2 ) f ( x1 )) xmax ( f ( xn
) f ( xn
1)) x max.
Then
U L ( f ( xn
) f ( x0 )) xmax ( f ( b) f ( a)) xmax f ( b) f ( a)
x max since f ( b) f ( a ). Thus
lim ( U L) lim ( f ( b) P 0 P 0
f ( a)) x max 0, since xmax P .
84. (a) U max1 x max2
x max n x where
max 1 f ( x0 ), max 2 f ( x1 ), ,max n f ( xn
1)
since f is decreasing on [ a, b];
L min1 x min2
x min n x where
min 1 f ( x1 ), min 2 f ( x2
), , min n f ( xn
)
since f is decreasing on [ a, b ]. Therefore
U L (max1 min 1) x (max2 min 2)
x
... (maxn min n)
x
( f ( x0 ) f ( x1 )) x ( f ( x1 ) f ( x2
)) x
... ( f ( xn
1) f ( xn
)) x
( f ( x0
) f ( xn
)) x ( f ( a) f ( b)) x.
(b) U max1 x max
1 2 x ... max
2
n x n where max 1 f ( x0 ), max 2 f ( x1 ), , max n f ( xn
1)
since f is decreasing on [ a, b];
L min1 x1 min2 x2
min n x n where
min f ( x ), min f ( x ), , min f ( x ) since f is decreasing on [ a, b ]. Therefore
1 1 2 2
U L (max min ) x (max min ) x (max min ) x
1 1 1 2 2 2
n
n
n n n
( f ( x ) f ( x )) x ( f ( x ) f ( x )) x ( f ( x ) f ( x )) x ( f ( x ) f ( x )) x
( f ( a) f ( b) xmax
f ( b) f ( a)
x max since f ( b) f ( a ). Thus
lim ( U L) lim f ( b) f ( a) x 0, since xmax P .
P
0 1 1 1 2 2 n 1 n n 0 n max
0 P 0
max
85. (a) Partition 0, into n subintervals, each of length x with points x
2
2n
0 0, x 1 x,
x 2 x,... , x n x . Since sin x is increasing on 0, , the upper sum U is the sum of the areas
2 n 2
of the circumscribed rectangles of areas f ( x1 ) x (sin x) x, f ( x2
) x (sin 2 x) x,... ,
f ( xn
) x (sin n x) x.
2
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