The tangent to the graph of the function $y = f(x)$ at the point with abscissa $x = a$ forms with the $x$-axis an angle of $\pi/3$ and at the point with abscissa $x = b$ at an angle of $\pi/4$. Then the value of the integral $\int_{a}^{b} f(x) \cdot f''(x) \, dx$ is equal to (assume $f''(x)$ to be continuous).

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 $A_n = \int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^n x \, dx$,then $\frac{A_4 - A_6}{A_4} = $

The value of $\int\limits_0^2 {\frac{{dx}}{{{{(1 - x)}^2}}}} $ is

If $f(x) = \int_0^x {t(\sin x - \sin t) dt}$,then which of the following is true?

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 $\left(0, \frac{\pi}{3}\right)$
$(C)$ $\lim_{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{1 - e^{x^2}} = -1$
$(D)$ $\lim_{x \rightarrow 0} \frac{\sin x \int_0^x f(t) dt}{x^2} = -1$