$\int_{-2}^2 (4-x^2)^{\frac{5}{2}} dx = $ (in $\pi$)

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

Let $f:(0, \infty) \rightarrow \mathbb{R}$ be given by $f(x)=\int_{\frac{1}{x}}^x e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$. Then
$(A)$ $f(x)$ is monotonically increasing on $[1, \infty)$
$(B)$ $f(x)$ is monotonically decreasing on $(0,1)$
$(C)$ $f(x)+f\left(\frac{1}{x}\right)=0$,for all $x \in(0, \infty)$
$(D)$ $f\left(2^x\right)$ is an odd function of $x$ on $\mathbb{R}$

If $\int_0^{t^2} xf(x)dx = \frac{2}{5}t^5, t > 0,$ then $f\left( \frac{4}{25} \right) = $

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

If $f(x) = \int_{9x^2}^{x^4} 5^{\sqrt{t}} dt$,then $\lim_{h \to 0} \frac{f(3 + h) - f(3 - h)}{h}$ is equal to