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:

If for a continuous function $f(x),$ $\int_{-\pi}^{t} (f(x) + x) dx = \pi^2 - t^2$ for all $t \ge -\pi,$ then $f\left(-\frac{\pi}{3}\right)$ is equal to

Let $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$,then:

Let $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cos^2 x \end{array} \right|$,then $\int_0^{\pi /2} f(x) dx = $

The value of $\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x}$ is

The angle of intersection between the curves $y = \int\limits_{x^2}^{x^3} \sqrt{5 - t^2} \, dt$ and the $x$-axis is (where $x \neq 0$):