power of integral I don't know how to put this properly but
Is there any some kind of equality or inequality like in case we know that $|\int_a^b f| \leq \int_a^b |f|$
$\Big[\int_a^b f(x) \Big]^{T}$ where $T \in \mathbb{R}^{+}$( positive real number) and $f(x)$ is positive for all $x$
 A: Jensen's inequality says that if $\varphi$ is convex then
$$ \varphi\left( \frac{1}{b-a} \int_a^b f \right) \le \frac1{b-a} \int_a^b \varphi \circ f.$$
The function $\varphi(x) = x^T$ is convex provided $T \ge 1$ so we have
$$ \frac{1}{(b-a)^T} \left( \int_a^b f \right)^T \le \frac1{b - a} \int_a^b f^T, \quad 1 \le T. $$
Which when we rearrange is
$$ \left( \int_a^b f \right)^T \le (b - a)^{T - 1} \int_a^b f^T, \quad 1 \le T. $$
If $b - a \le 1$ then this implies the weaker inequality
$$ \left( \int_a^b f \right)^T \le \int_a^b f^T, \quad 1 \le T. $$
Also, if $0 \le T \le 1$ the function is concave so we have the opposite inequality:
$$ \frac{1}{(b - a)^T} \left( \int_a^b f \right)^T \ge \frac1{b- a} \int_a^b f^T, \quad 0 < T \le 1. $$
A: In general the statement is not true. A counter-example is found by taking $a = 0, b= 15, f(x) = x^2$ and $T = 2$, since then
\begin{align*}
\mathopen{}\left[\int_{a}^{b}f(x)\,\mathrm{d}x \right]^{T}\mathclose{} &= \mathopen{}\left[\int_{0}^{15}x^{2}\,\mathrm{d}x\right]^{2}\mathclose{} \\[1ex]
&=1\,265\,625
\end{align*}
while
\begin{align*}
\int_{a}^{b}\Big[ f(x)\Big]^{T}\,\mathrm{d}x  &= \int_{0}^{15}x^{4}\,\mathrm{d}x \\[1ex]
&=151\,875.
\end{align*}
