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44 Thefunction f is continuous on [0, 1] by qn 6.49 and is thus integrable.
s ( x )
1 is an upper step
function for g . Consider step functions on [0, 1/ n ] and [1/ n , 1]. The
function is continuous on [1/ n , 1] and therefore there are arbitrarily
close upper and lower sums. On [0, 1/ n ] the difference between upper
and lower sums is 2/ n which may bearbitrarily small. So thefunction
is integrable by qn 24.
1 is a lower step function and S ( x )
45 See qns 16 and 18. Since f is continuous, ( b
a ) f is continuous. The
minimum valueof this function is m ( b
a ) and themaximum valueof
this function is M ( b
a ). Since the integral lies between these values,
or at one of them, the integral is equal to ( b
a ) f ( c ) for some c
[ a , b ]
by the Intermediate Value Theorem.
Dividethersult of qn 43(ii) by ( b a ).
46 Since f is integrable on [ a , b ], there exist step functions s and S giving
upper and lower sums which arearbitrarily clos. If c and d were not
part of the subdivisions for these step functions, introduce these points
into the subdivision for each step function. Then the difference between
the upper and lower sums on [ c , d ]
the difference between the upper
and lower sums on [ a , b ], so f is integrable on [ c , d ].
47 If f is a step function the result is obvious. To obtain the result for any
integrable f , apply this result to upper and lower step functions for f .
f
48 To retain the equation for qn 47, we must define
0 and
f
f .
49 Thefunction f is monotonic.
0
when 0
x
1,
x
x
F ( x )
1
when 1
2,
2 x
x
3
when 2
3.
F is continuous at x 1 and 2, by considering limits from above
and below at each of these points.
50 Since f is integrable, f is bounded. Let m f ( x )
M .
F ( c h )
F ( c )
f , so for positive h ,
mh F ( c h )
F ( c )
Mh .
( F ( c h )
F ( c ))
F ( c h )
F ( c ).
Thus lim
0, so lim
Likewise for the limit from below. Then use qn 6.89, continuity by
limits.
L max m , M .
51
(i) By the Mean Value Theorem applied to F .
(iii) Every summation of this kind lies between an upper sum and
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