It says if f(x) => g(x) for a<t<b explain briefly why the integral of f(x) => integral of g(x) in the range a to b.
Then it says if p>q>0 and x>= 1 then prove
1/p*(x^p -1) => 1/q(x^q - 1)
and show this also holds when p>q>0 and 0<=x<=1.
then lastly it says prove that if p>q>0 and x=>0, then
1/p*(x^p/(p+1) -1) => 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)>t(q+1).
Is this correct?
Then for the p>q>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.