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Unformatted text preview: unction on an interval a, b , that g is nonnegative and integrable on a, b .
Prove that, if m and M are, respectively, the minimum and maximum values of f, then
mÞ g Þ a fg b a and deduce that there exists a number c MÞ g b b a a, b such that Þa f g f Þ a g. b b c This fact is sometimes called the mean value theorem for integrals.
Since
m fx
M
for every x
a, b we have Þ a mg Þ a fg Þ a Mg
b b b which gives us
mÞ g Þ a fg MÞ g Þ a fg b M. b a b a and so
b m Þa g
b The existence of the number c follows at once from the Bolzano intermediate value theorem.
b. Given that f is continuous on an interval a, b , prove that there exits a number c Þa f f c a, b such that b 6. b a. a. Given that f is a nonnegative continuous function on an interval a, b where a b and that Þ f 0,
a
prove that f is the constant zero function.
To obtain a contradiction, suppose that c
a, b and that f c 0. Using the fact that f is
continuous at c, choose 0 such that the inequality
fc
fx
2
holds for every number x
a, b
c , c . The set a, b
c , c is an interval with
positive length that we shall write as c, d . In fact, c is the larger of the two numbers a and c
and d is the smaller of the two numbers b and c . Since
b
d
d
c
c
Þa f Þc f Þc f 2 f 2 d c 0
which gives us our desired contradiction.
b b. Given that f is a continuous function on an interval a, b where a b and that Þ f 0, for every
a
x
a, b , prove that f is the constant zero function.
We know that if
x Fx Þa f
x for each x then F x f x for each x. Since F is the constant function zero we conclude that
f x 0 for each x.
7. In this exercise we consider another proof of the “u decreasing” form of the change of variable theorem.
a. Given that f is an integrable function on an interval a, b and that g t f t whenever b 279 t a, give a direct proof that g is integrable on the interval b, a and that Þ b g t dt Þ a f x dx.
a b Suppose first that f is a step function on a, b that steps within the partition
P x 0 , x 1 , , x n
taking the constant value j on each interval x j 1 , x j . We define
Q x n , x n 1 , , x 1 , x 0
and observe that Q is a partition of the interval b, a . Now we define g t f t for each
t
a, b and we observe that g is a step function that takes the value j on each interval
x j , x j 1 . We see at once that Þa f n b n j xj xj 1 xj j1 xj j 1 j1 Þ bg
a We can now handle the general case. Using the fact that f is integrable on the interval a, b ,
choose a pair of sequences of step functions that squeezes f on the interval b, a . In other words,
sn f Sn
for each n and Þa Sn
b lim nÝ s n 0. For each n we define s n t s n t and S n t S n t and we observe that
sn g Sn
and, by the case we have already considered we deduce that Þ b Sn
a as n Þa Sn
b sn Ý. It follows that the function g is integrable on the interval
lim
Þ b g n Ý Þ b sn
a 1. 0 sn a nÝ
lim b, a and that Þa sn Þa f .
b b a. Suppose that u is a decreasing diffe...
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 Fall '08
 STAFF
 Math, Calculus

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