If $\varphi (x) = \int_{1/x}^{\sqrt{x}} \sin(t^2) \, dt$,then $\varphi'(1) = $

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

$\int_0^2 x^{\frac{5}{2}} \sqrt{2-x} \, dx =$

$\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx = $

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 $.........$

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: