Let $f(\alpha) = \int_{0}^{\alpha} x^{2} \left(1 - \frac{x}{\alpha}\right)^{\alpha} dx$ (where $\alpha > 0$),then $\sum_{\alpha=1}^{5} \frac{f(\alpha)}{\alpha^{3}}$ is equal to-

Let $f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ be a continuous function such that $f(0)=1$ and $\int_0^{\frac{\pi}{3}} f(t) dt = 0$. Then which of the following statements is (are) $TRUE$?
$(A)$ The equation $f(x) - 3 \cos 3x = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
$(B)$ The equation $f(x) - 3 \sin 3x = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
$(C)$ $\lim_{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{1 - e^{x^2}} = -1$
$(D)$ $\lim_{x \rightarrow 0} \frac{\sin x \int_0^x f(t) dt}{x^2} = -1$

Let $I_1 = \int_0^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx$,$I_2 = \int_0^{2\pi} \cos^6 x dx$,$I_3 = \int_{-\pi/2}^{\pi/2} \sin^3 x dx$,and $I_4 = \int_0^1 \ln \left( \frac{1}{x} - 1 \right) dx$. Then:

If $f(x) = \text{Max}\{\sin x, \cos x\}$ and $g(x) = \text{Min}\{\sin x, \cos x\}$,then $\int_{0}^{\pi} f(x) dx + \int_{0}^{\pi} g(x) dx = $

Let $f: R \rightarrow R$ be a differentiable function such that its derivative $f^{\prime}$ is continuous and $f(\pi)=-6$. If $F:[0, \pi] \rightarrow R$ is defined by $F(x)=\int_0^{ x } f( t ) dt$,and if $\int_0^\pi\left(f^{\prime}( x )+ F ( x )\right) \cos x dx =2$,then the value of $f(0)$ is.

If $\alpha \in (2, 3)$,then the number of solutions of the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$ is: