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

If $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$ for all $x \in R - \{0\}$,then the value of $f(\frac{\pi}{6})$ is

$\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\} = $

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\int_0^x {\cos {t^2}dt} }}{x}$ is

$\int_0^{\pi/6} \cos^4 3\theta \cdot \sin^2 6\theta \, d\theta$ is equal to

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