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

Investigate for maxima & minima for the function, $\text{f}\left(\text{x}\right)$, $\text{f}\left(\text{x}\right)=\underset{1}{\overset{\text{x}}{\int }}\left[2\left(\text{t}-1\right){\left(\text{t}-2\right)}^3+3{\left(\text{t}-1\right)}^2{\left(\text{t}-2\right)}^2\right]\text{dt}$

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 $f:\left(0, 1\right)\to \mathrm{ℝ}$ be the function defined as $f\left(x\right)=\sqrt{n}$ if $x\in [\frac{1}{n+1}, \frac{1}{n})$ where $n\in \mathrm{\mathbb{N}}$. Let $g:\left(0, 1\right)\to \mathrm{ℝ}$ be a function such that ${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g\left(x\right)<2\sqrt{x}$ for all $x\in \left(0, 1\right)$. Then $\underset{x\to 0}{\lim }f\left(x\right) g\left(x\right)$

Let f be a continuous function satisfying ${\int }_0^{{\mathrm{t}}^2}\left(\mathrm{f}\left(\mathrm{x}\right)+{\mathrm{x}}^2\right)\mathrm{dx}=\frac{4}{3}{\mathrm{t}}^3,\forall \mathrm{t}>0$ . Then $\mathrm{f}\left(\frac{{\mathrm{\pi }}^2}{4}\right)$ is equal to

If ${\int }_0^{t^2}\left(f\left(x\right)+x^2\right)dx=\frac{4}{3}{t}^3$, then $f\left(x\right)$ is

Let $f$ be a differentiable function satisfying ${\int }_0^x\left(x-p+1\right)f\left(p\right)dp=x^4+x^2 \forall x\in [-2,\infty )$, then value of ${\int }_{-2}^2\left(f\left(x\right)+f^{\text{'}}\left(x\right)\right)dx$ is