If $f(x) = \int_{\pi^2/16}^{x^2} \frac{\sin x \cdot \sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta$,then the value of $f'(\frac{\pi}{2})$ is:

Let $F(x) = \int_x^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt$ for all $x \in \mathbb{R}$ and $f: [0, \frac{1}{2}] \rightarrow [0, \infty)$ be a continuous function. For $a \in [0, \frac{1}{2}]$,if $F'(a) + 2$ is the area of the region bounded by $x=0, y=0, y=f(x)$ and $x=a$,then $f(0)$ is

The minimum value of the twice differentiable function $f(x) = \int_{0}^{x} e^{x-t} f'(t) dt - (x^2 - x + 1) e^x, x \in R$,is.

Let $f(x) = \int_{\sin x}^{\cos x} e^{-t^2} dt$. Then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals

Given that $\frac{d}{d x}\left[\int_0^{\phi(x)} f(t) d t\right]=f(\phi(x)) \cdot \phi^{\prime}(x)$. If $\int_0^{x^3} f(t) d t = x^2 \sin(2 \pi x)$,then the value of $f(8)$ is

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 $.........$