If $f(x) = \int_a^x {t^3 e^t \, dt}$,then $\frac{d}{dx} 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

The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$ is equal to

The points of extremum of $\int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$ are

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

$\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)+\sin ^5\left(\frac{3 \pi}{6 n}\right)+\ldots+\sin ^5\left(\frac{\pi}{2}\right)\right\} = $