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

Let $f$ be a twice differentiable function on $\mathbb{R}$. If $f^{\prime}(0)=4$ and $f(x)+\int_{0}^{x}(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$,then $(2 a+1)^{5} a^{2}$ is equal to $\dots\dots$

Let $f : (-1, 1) \to R$ be a continuous function. If $\int\limits_0^{\sin x} {f(t)dt} = \frac{\sqrt{3}}{2}x$,then $f\left(\frac{\sqrt{3}}{2}\right)$ is equal to

$\int_{-2}^2 x^4(4-x^2)^{\frac{7}{2}} dx=$

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 $S(x) = \int_{x^2}^{x^3} \ln t \, dt$ for $x > 0$ and $H(x) = \frac{S'(x)}{x}$. Then $H(x)$ is :