# Math Help - Integral Inequality

1. ## Integral Inequality

Hi,

I have the following inequality
$\int_0^t \! \, F(\tau) d\tau \leq M$
where $F(\tau)>0$ is a continuous differentiable function and $M$ is a constant.

My question, what is the upper bound for
$\int_0^t \! \, F^n(\tau) d\tau \leq ?$
where $n$ is positive integer larger then 2.

2. There is no upper bound. Choose a number c such that 0<c<t, let f(x) = M/c when x < c and 0 otherwise. f is not continuous differentiable but one is always able to find a differetiable function F as close as possible to f. Now $\int_0^t f(x)dx = M$, and $\int_0^t f^n(x)dx=c(\frac{M}{c})^n=\frac{M^n}{c^{n-1}}$. Choose a small c you can get an arbitary big result.

3. Thank you for your reply. How if t>1, if I choose a sufficiently large c, 1<c<t, then the bound will get smaller at t increase.