Unformatted text preview: rentiable function on an interval a, b and that the derivative u of u is
integrable on a, b . Apply the form of the monotone version of the change of variable theorem
proved above to the function v defined by the equation v t u t for b t
a to show that the
equation Þa f u t
b u t dt Þu a ub f x dx holds for every function f integrable on the interval u b , u a .
We define g t f t for all t
a, b . Then g is integrable on Þ ua ub g Þu b ua Þu a ua, ub ub f f From the monotone version of the theorem proved earlier we also see that Þ ua g Þv a g ub Þa g ut vb Þa g v t
b b v t dt u t dt Þa f u t
b We conclude that Þa f u t
b 280 u t dt Þu a ub f u t dt. and we have To reach some additional exercises that invite you to develop some important inequalities, click on the icon
. Additional Exercises on the Change of Variable
The exercises in this document invite you to derive some inequalities that have important applications in
mathematics and other disciplines.
1. Given that f is a strictly increasing continuous function on an interval a, b , prove that Þa f Þf a
b fb f 1 bf b af a . Solution:
a. First we shall motivate this problem by looking at the special case in which f has a continuous
derivative. In this case we can change variable in the integral Þf a fb f 1 to obtain Þf a fb f 1 Þa f 1 f t
b f t dt Þ a tf
b t dt which we can integrate by parts to obtain Þ a tf
b Þ a 1f t dt.
b t dt bf b af a So the desired identity is true in this special case. Incidentally, this special case is all that we need for
the application of the identity to Exercise 3.
b. Now we consider the general case. The proof in this case is not difficult but it depends upon
Darboux’s theorem. (As a matter of fact, all the proof needs is the application of Darboux’s theorem
to continuous functions and this form of Darboux’s theorem is easier to prove.) Suppose that
P x 0 , x 1 , , x n
is a the regular n-partition of the interval a, b for each n and, for each j, suppose that y j f x j .
Then if Q n is the partition
Q n y 0 , y 1 , , y n
of the interval f a , f b then, since f is uniformly continuous on a, b , the mesh of the partition Q n
must approach 0 as n Ý. Now for each n we see that
n n xj
j1 xj 1 f xj n yj yj 1 f 1 yj 1 j1 n xj xj 1 yj j1 yj yj 1 xj 1 j1 n xj xj 1 yj yj yj 1 xj 1 j1
n xjyj yj 1xj j1 bf b
and so, letting n Ý, we see that 281 af a 1 xnyn x0y0 Þa f Þf a
b fb f bf b 1 af a . 2. Given that f is a strictly increasing continuous unbounded function on the interval 0, Ý and that f 0 0,
prove that for all positive numbers a and b we have Þ0 f Þ0 f 1.
a ab b Solution: The Case b f a :
The case b f a is illustrated in the following figure.
(a,f(a)) (0,0) The case b f a
The area of the yellow region is Þ f 1 and the area of the turquoise region is Þ f. The sum of these two
areas is ab. Now we say it precisely: It follows from the preceding exercise that
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