Let $f$ be a real-valued continuous function on $[0, 1]$ and $f(x) = x + \int_{0}^{1} (x - t) f(t) dt$. Then which of the following points $(x, y)$ lies on the curve $y = f(x)$?

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

The value of $\int\limits_0^{\frac{\pi }{2}} {\sin 8x \cot x \, dx} + \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\ln \left( {\frac{{1 - \sin x}}{{1 + \sin x}}} \right)dx}$ is equal to

For $x, t \in R$,let $p_t(x) = (\sin t) x^2 - (2 \cos t) x + \sin t$ be a family of quadratic polynomials in $x$ with variable coefficients. Let $A(t) = \int_0^1 p_t(x) dx$. Which of the following statements are true?
$I$. $A(t) < 0$ for all $t$.
$II$. $A(t)$ has infinitely many critical points.
$III$. $A(t) = 0$ for infinitely many $t$.
$IV$. $A'(t) < 0$ for all $t$.

If for a real number $y$,$[y]$ is the greatest integer less than or equal to $y$,then the value of the integral $\int_{\pi /2}^{3\pi /2} [2\sin x] \, dx$ is

The function $f(x) = \int\limits_0^x \sqrt{1 - t^4} \, dt$ is such that