$\int_9^x \frac{f(y)}{y^2} \, dy = 2 \sqrt{x} - 6 \implies f(x) = ?$

Let $f, g:(0, \infty) \rightarrow \mathbb{R}$ be two functions defined by $f(x)=\int_{-x}^x(|t|-t^2) e^{-t^2} dt$ and $g(x)=\int_0^{x^2} t^{1/2} e^{-t} dt$. Then the value of $(f(\sqrt{\log_{e} 9}) + g(\sqrt{\log_{e} 9}))$ is

Let $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$ where $x > 0$ and $f$ is a continuous function such that $f(2x) = f(x)$. Then:

If $f(x) = \int_{x^2}^{x^4} \sin \sqrt{t} \, dt$,then $f'(x)$ equals

$\int_0^{\pi /2} \sin^{2m} x \, dx = $

Let $f : R \rightarrow R$ be a continuous odd function,which vanishes exactly at one point and $f(1) = \frac{1}{2}$. Suppose that $F(x) = \int_{-1}^x f(t) dt$ for all $x \in [-1, 2]$ and $G(x) = \int_{-1}^x t|f(f(t))| dt$ for all $x \in [-1, 2]$. If $\lim_{x \rightarrow 1} \frac{F(x)}{G(x)} = \frac{1}{14}$,then the value of $f\left(\frac{1}{2}\right)$ is