I apologize for not knowing how to use Latex, so I will type the problem as it is read...
Prove that for all x greater than or equal to 0, we have
the integral from 0 to x for [g(x)]^3 dx which is less than or equal to
(the integral from 0 to x for g(x) dx)^2.
Suppose that g is a differentiable function with g(0) = 0 and 0 is less than g prime which is less than or equal to 1.
Also, the brackets around the g(x) are supposed to be there, so it is to the power 3, as in C^3. The parentheses around the second part of the problem are there to show it is to the power 2.
Sorry for the non-Latex format.
I do not know where to start, but think it involves MVT (the Mean Value Theorem) and/or Rolle's Theorem.
The Attempt at a Solution
I tried working with a square root, but was unsure. My skills are rusty.
Thanks for any help.