Newton-Leibnitz's Formula

Let $f$ be a real-valued function defined on the interval $\left(-1, 1\right)$ such that $e^{-x}f\left(x\right)=2+{\int }_0^x\sqrt{t^4+1}dt$, for all $x\in \left(-1, 1\right)$ and let $f^{-1}$ be the inverse of $f$. Then, ${\left(f^{-1}\right)}^{\text{'}}\left(2\right)$ is equal to

If $f\left(x\right)={\int }_0^x{\left\{f\left(t\right)\right\}}^{-1}dt$ and ${\int }_0^1{\left\{f\left(t\right)\right\}}^{-1}dt=\sqrt{2}$, then

If ${\int }_0^x\frac{bt\cos 4t-a\sin 4t}{t^2}dt=\frac{a\sin 4x}{x}$ for all $x\ne 0$, then $a$ and $b$ are given by

 The value of $\underset{x\to 0}{\lim }\frac{2{\int }_1^{\cos x}{\cos }^{-1}\left(t\right)}{2x-\sin 2x}dt$ is

Let $f\left(x\right)={\int }_1^x\sqrt{2-t^2}dt$, then the real roots of the equation $x^2-f^{\text{'}}\left(x\right)=0$ are

Let $f : (0, \infty )\to R$ and $F\left(x\right)={\int }_0^{x}f\left(t\right)dt.$ If $F\left(x^2\right)=x^2(1+x)$, then $f\left(4\right)$ is equal to

The points of extremum of $\varnothing \left(\mathrm{x}\right)={\int }_1^{\mathrm{x}}{\mathrm{e}}^{-{\mathrm{t}}^2/2} (1-{\mathrm{t}}^2)\mathrm{dt}$ are

If ${\int }_0^{x}f\left(t\right)dt=x+{\int }_x^{1}tf\left(t\right)dt$, then the value of $f\left(1\right)$ is

For $x\in \left(0,\frac{5\pi }{2}\right)$, define $f\left(x\right)=\underset{0}{\overset{x}{\int }}\sqrt{t}\sin tdt$. Then, $f$ has

Let $f:\left[0, 2\right]\to R$ be a function which is continuous on $\left[0,2\right]$ and is differentiable on $\left(0,2\right)$ with $f\left(0\right)=1.$ Let $F\left(x\right)=\underset{0}{\overset{x^2}{\int }}f\left(\sqrt{t}\right)dt$, for $x\in \left[0,2\right]$. If $F^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right),\forall x\in \left(0,2\right)$, then $F\left(2\right)$ equals

Let $f\left(x\right)=\underset{0}{\overset{g\left(x\right)}{\int }}\frac{dt}{\sqrt{1+t^2}}$, where $g\left(x\right)=\underset{0}{\overset{\cos x}{\int }}\left(1+\sin t^2\right)dt.$ Also, $h\left(x\right)=e^{-\left|x\right|}$ and $f\left(x\right)=x^2\sin \frac{1}{x},$ if $x\ne 0$ and $f\left(0\right)=0$, then $f^{\text{'}}\left(\frac{\mathrm{\pi }}{2}\right)$ is equal to

Let $f\left(x\right)=\underset{-1}{\overset{x}{\int }}{e}^{t^2}dt$ and $h\left(x\right)=f\left(1+g\left(x\right)\right)$, where $g\left(x\right)$ is defined for all $x, g^{\text{'}}\left(x\right)$ exists for all $x,$ and $g\left(x\right)<0$ for $x>0.$ If $h^{\text{'}}\left(1\right)=e$ and $g^{\text{'}}\left(1\right)=1$, then the possible values which $g\left(1\right)$ can take

If $a, b$ and $c$ are real numbers, then the value of $\underset{t\to 0}{\lim }\ln \left(\frac{1}{t}{\int }_0^t{\left(1+a\sin bx\right)}^{\frac{c}{x}}dx\right)$ equals

If $f\left(x\right)=e^{g\left(x\right)}$ and $g\left(x\right)={\int }_2^x\frac{tdt}{1+t^4},$ then $f^{\text{'}}\left(2\right)$ is equal to

Let $C_n=\underset{\frac{1}{n+1}}{\overset{\frac{1}{n}}{\int }}\frac{{\tan }^{-1}\left(nx\right)}{{\sin }^{-1}\left(nx\right)}dx$, then $\underset{n\to \infty }{\lim }{n}^2·C_n$ is equal to

If $\underset{\frac{\mathrm{\pi }}{2}}{\overset{x}{\int }}\sqrt{\left(3-2{\sin }^{2}t\right)}dt+\underset{0}{\overset{y}{\int }}\cos tdt=0$, then $\left(\frac{dy}{dx}\right)$ at $x=\mathrm{\pi }$ and $y=\mathrm{\pi }$ is

Let $f$ be a non-negative function defined on the interval $\left[0,1\right].$  If $\underset{0}{\overset{x}{\int }}\sqrt{1-{\left(f^{\text{'}}\left(t\right)\right)}^2}dt =\underset{0}{\overset{x}{\int }}f\left(t\right)dt,0,x\le 1$ and $f\left(0\right)=0,$  then