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

Let $f$ and $g$ be continuous, real valued functions on the closed interval $[a, b],$ and $F$ and $G$ be functions such that $F\text{'}\left(x\right)=f\left(x\right)$ and $G\left(x\right)={\int }_a^{x}g\left(t\right)dt$ for $x$ in $[a, b].$ Then

Suppose $f : \mathrm{ℝ}\to \mathrm{ℝ}$ is a continuous function and $F:\mathrm{ℝ}\to \mathrm{ℝ}$ is defined by $F\left(x\right)={\int }_0^{x}f\left(t\right)\sin (x-t)dt$ for $x\in \mathrm{ℝ}$ . Then

If $f\left(x\right)={\int }_0^{x^2}\frac{dt}{1+t^3}$, then $f^{\text{'}}\left(2\right)$ is -

Let $f:\left[0,\frac{\pi }{2}\right]\to R$ be continuous and satisfy $\underset{0}{\overset{\sin x}{\int }}f\left(t\right)dt=\sqrt{3}x/2$ for $0\le x\le \frac{\pi }{2}$. Then $f\left(\frac{1}{2}\right)$ equals -