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Topic: Inequality help!
Replies: 4   Last Post: Dec 4, 2004 8:37 PM

 Messages: [ Previous | Next ]
 steve Posts: 32 Registered: 12/6/04
Inequality help!
Posted: Nov 23, 2004 4:49 PM

Hi I am having trouble with this question I saw in a problem solving

I know its hard to read inequalities nicely so I have also put a
picture of the question on the internet here.

<a href="http://img44.exs.cx/img44/2759/q24.jpg">http://img44.exs.cx/img44/2759/q24.jpg</a>

It says if f(x) =&gt; g(x) for a&lt;t&lt;b explain briefly why the integral of
f(x) =&gt; integral of g(x) in the range a to b.

Then it says if p&gt;q&gt;0 and x&gt;= 1 then prove

1/p*(x^p -1) =&gt; 1/q(x^q - 1)

and show this also holds when p&gt;q&gt;0 and 0&lt;=x&lt;=1.

then lastly it says prove that if p&gt;q&gt;0 and x=&gt;0, then

1/p*(x^p/(p+1) -1) =&gt; 1/q(x^q/(q+1) - 1)

this is what I have done so far.

Really briefly this is to do with every value of f(x) is bigger then
or equal to g(x) so if we take the integral to mean the area under the
graph then it is clear the integral of f(x) is bigger then g(x) as it
is the same length under the graph but every value is more.

Then I am not sure if this is correct but for part 2,

if we consider the integral from 1 to x of t^(p+1) and same integral
of t^(q+1), it gives us the result from the first part. as
t^(p+1)&gt;t(q+1).

Is this correct?

Then for the p&gt;q&gt;0, if we multiply the integral in part 1 by -1 then
change the 1 and the x around in the integral we get the desired
result (I hope).

But then the last part I have no idea what to do, could someone please
help.

thank you

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Date Subject Author
11/23/04 steve
11/24/04 John
11/24/04 Darrell
11/24/04 steve
12/4/04 steve