Alright, I won't kid around. This problem requires a solution and it has been causing my head some trouble. This is a problem from the Cambridge STEP specimen paper and it goes like this:

Prove that (x^p -1)/p is greater than or equal to (x^q -1)/q.
Where p> q> 0 and 0 < x < 1.

You will have seen the first part of the question which is to explain why ${\int }_a^{b}f(x)\mathrm{dx}>{\int }_a^{b}g(x)\mathrm{dx}$ if $a<b$ and $f(x)>g(x)$ for all $x$ in $(a,b)$ - this is clear from areas.

Note that $(x^p-1)/p={\int }_1^x{t}^p\mathrm{dt}$.

For $0<x<1$, $x^p<x^q$, so ${\int }_x^1{t}^p\mathrm{dt}<{\int }_x^1{t}^p\mathrm{dt}$.

For these integrals the limits are "the right way round" since x < 1.

But ${\int }_1^x{t}^p\mathrm{dt}=-{\int }_x^1{t}^p\mathrm{dt}$. Thus the required inequality follows.

If this doesn't make sense, sorry - let me know and I'll write it out a bit more clearly.

David

Thanks for the explanation David. However, I guess that I find it difficult to read. Using your method ( or what I can make out of it ), the inequality I end up with gives me an incorrect answer, ( ie:the opposite to what the STEP paper says it should be ). So I would be glad to take you up on your offer of a fuller explanation.

David,

Your method is fine. Just be careful since the question asks for p> q> 0, not p < q, so we obtain the required inequality.

Kerwin