Definite Integration

Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x}f(x)=2+\displaystyle\int_0^x\sqrt{t^4+1}\,dt$, for all $x\in(-1,1)$, and let $f^{-1}$ be the inverse function of $f$. Then $(f^{-1})'(2)$ is equal to

Match the statements in Column-I with the values in Column-II.

Let $f$ be a non-negative function defined on the interval $[0,1]$. If
$$\int_0^x\sqrt{1-(f'(t))^2}\,dt=\int_0^xf(t)\,dt,\qquad0\le x\le1,$$
and $f(0)=0$, then

Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is

Match the statements/expressions in Column I with the open intervals in Column II.

If
$$I_n=\int_{-\pi}^{\pi}\frac{\sin nx}{(1+\pi^x)\sin x}\,dx,\qquad n=0,1,2,\ldots,$$
then

Let $f:\mathbf R\to\mathbf R$ be a continuous function which satisfies
$$f(x)=\int_0^xf(t)\,dt.$$
Then the value of $f(\ln5)$ is

Let
$$S_n=\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}\quad\text{and}\quad T_n=\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2},$$
for $n=1,2,3,\ldots$. Then,

Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty,\infty)$ such that $f(x)=f(1-x)$ and $f'\left(\frac14\right)=0$. Then,

$\displaystyle\int_{-1}^1g'(x)\,dx=$

Let
$$g(x)=\int_0^{e^x}\frac{f'(t)}{1+t^2}dt.$$
Which of the following is true?

$$\lim_{x\to\frac\pi4}\frac{\displaystyle\int_2^{\sec^2x}f(t)\,dt}{x^2-\frac{\pi^2}{16}}$$
equals

Match the integrals in Column I with the values in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.