If $F(x) = \int_{x^2}^{x^3} \log t \, dt$ $(x > 0)$,then $F'(x) = $

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

If $x \cdot \sin(\pi x) = \int_{0}^{x^2} f(t) \, dt$ where $f$ is a continuous function,then the value of $f(4)$ is:

If $\int_{\sin x}^1 {{t^2}f(t)\;dt = 1 - \sin x} $,$x \in \left( {0,\frac{\pi }{2}} \right)$,then $f\left( {\frac{1}{{\sqrt 3 }}} \right)$ is equal to:

Let $f$ be a differentiable function satisfying $f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left(\frac{\lambda^{2} x}{3}\right) d\lambda$ for $x > 0$ and $f(1) = \sqrt{3}$. If $y = f(x)$ passes through the point $(\alpha, 6)$,then $\alpha$ is equal to $.........$