Newton Leibnitz's Theorem

Let $f\left(x\right)=\underset{1}{\overset{x}{\int }}\sqrt{2-t^2}dt$. Then the real roots of the equation $x^2-f^{\text{'}}\left(x\right)=0$ are

If $y=\frac{1}{a}\underset{0}{\overset{x}{\int }}f\left(t\right)·\sin a\left(x-t\right)dt$, then find$f\left(x\right)$

If $\mathrm{\varphi }\left(x\right)=\cos x-\underset{0}{\overset{x}{\int }}\left(x-t\right)\mathrm{ \varphi }\left(t\right)dt.$ Then find the value of ${\varphi }^{\text{'}\text{'}}\left(x\right)+\varphi \left(x\right).$

$f\left(x\right)=\sin \mathit{x}+{\int }_0^x{f}^{\text{'}}\left(t\right)\left(2\sin \mathit{t}-{\sin }^{2}t\right)dt$, then $f\left(x\right)$ is

Let $f\left(x\right)={\int }_{-x}^x\left(t\sin at+bt+c\right)dt$, where $a,b,c$ are non-zero real numbers, then $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}$ is

The value of  $\underset{x\to \infty }{\lim }\left(x^3{\int }_{-\frac{1}{x}}^{\frac{1}{x}}\frac{\ln \left(1+t^2\right)}{1+e^t}dt\right)$ is

Let $f\left(x\right)={\int }_2^x\frac{dt}{\sqrt{1+t^4}}$ and $g$ be the inverse of $f$. Then the value of $g^{\text{'}}\left(0\right)$ is

The value of $\underset{x\to 0}{\lim }\frac{1}{x^3}\underset{0}{\overset{x}{\int }}\frac{t\ell n(1+t)}{t^4+4}dt$ is

The value of  $\underset{x\to 0}{\lim } \frac{1}{x^3}\underset{0}{\overset{x}{\int }}\frac{t \ln \left(1+t\right)}{t^4+4}dt$  is:

If $f\left(x\right)$ is differentiable and ${\int }_0^{t^2}xf\left(x\right)dx=\frac{2}{5}{t}^5,$then $f\left(\frac{4}{25}\right)$ equals to

If $f\left(x\right)$ is differentiable and $\underset{0}{\overset{t^2}{\int }}xf\left(x\right)dx=\frac{2}{5}{t}^5,$ then $f\left(\frac{4}{25}\right)$ equals:

Let $f\left(x\right)=\underset{1}{\overset{x}{\int }}\sqrt{2-t^2}dt$.  Then the real roots of the equation $x^2-f^{\text{'}}\left(x\right)=0$ are

Let $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt,$ where $f$ is such that $\frac{1}{2}\le f\left(t\right)\le 1,$ for $t\in \left[0,1\right]$ and $0\le f\left(t\right)\le \frac{1}{2},$ for $t\in \left[1,2\right]$. Then $g\left(2\right)$ satisfies the inequality

Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt,$ where $f$ is such that  $\frac{1}{2}\le f\left(t\right)\le 1$ for $t\in [0,1]$ and $0\le f\left(t\right)\le \frac{1}{2}$ for $t\in [1,2].$ Then, $g\left(2\right)$ satisfies the inequality

If $f\left(x\right)={\int }_e^{e^x}{\log }_e\left(\frac{x}{{\log }_{e}t}\right)dt,$ then the value of $\frac{3f^{\text{'}}\left(3\right)}{e}$ is

Let $F\left(x\right)={\int }_0^{x}f\left(t\right)dt$, where $f\left(x\right)=2+\sin x-\cos x$. If $\left|F\left(x\right)-F\left(y\right)\right|\le k\left|x-y\right|$ for all $x$ and $y$ in $R$, then a possible value of $k$ is

If $y=\underset{1}{\overset{x^4}{\int }}\sin \sqrt{t}·dt$ , then the value of $\frac{dy}{dx}$ at $x=0$ is

If $\int f\left(x\right)\cos xdx=\frac{1}{2}{\left\{f\left(x\right)\right\}}^2+C$$,$ then $f\left(\frac{\pi }{2}\right)$ is, (where $C$ is an arbitrary constant)