Definite Integration

If $y=x^{\underset{1}{\overset{x}{\int }}\ln tdt}$, then the value of $\frac{dy}{dx}$ at $x=e$ is

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 ${\varphi }^{\text{'}\text{'}}\left(x\right)+\varphi \left(x\right).$

$I={\int }_0^{\pi }\frac{x^{2 }\sin 2x\sin \left(\frac{\pi }{2}\cos x\right)}{2x-\pi } dx$

If ${\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{\left({\sin }^3\theta -{\cos }^3\theta -{\cos }^2\theta \right){\left(\sin \theta +\cos \theta +{\cos }^2\theta \right)}^{2007}}{{\left(\sin \theta \right)}^{2009}{\left(\cos \theta \right)}^{2009}}d\theta$$=\frac{{\left(a+\sqrt{b}\right)}^d-{\left(1+\sqrt{c}\right)}^d}{d}$, where $a, b, c$  and $d$ are all positive integers. Then the value of $(a+b+c+d)$ is

The value of : $\underset{0}{\overset{2}{\int }}\sqrt{4-x^2}dx$ would be:

The value of $\underset{\mathrm{a}}{\overset{b}{\int }}f\left(x\right)dx=\left(b-a\right)\underset{n\to \infty }{\lim }\frac{1}{n}[f\left(a\right)+f\left(a+h\right)+\mathrm{...}+f\left(a+\left(n-1\right)h\right)],$ where $h=\frac{b-a}{n}$, $f\left(x\right)=x^2+x+2; a=0, b=2$ as limits of sum would be

The value of ${\int }_1^{e^{37}}\frac{\mathrm{\pi sin}\left(\mathrm{\pi }\ell nx\right)}{x}dx$ is

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

The value of the integral ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{\frac{\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}1-x}{1+x}}dx}$ is

The value of the integral $\underset{0}{\overset{1}{\int }}\sqrt{\frac{\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}1-x}{1+x}}dx$ is

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  $f\left(x\right)=x-\left[x\right],$  for every real number $x,$ where $\left[x\right]$ is the integral part of $x.$ Then  $\underset{-1}{\overset{1}{\int }}f\left(x\right)dx$ is:

If $f\left(x\right)=Asin\left(\frac{\pi x}{2}\right)+B, f\text{'}\left(\frac{1}{2}\right)=\sqrt{2} \mathrm{and}\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{2A}{\pi },$ then constants $A$ and $B$ are respectively 

$f\left(x\right)=A\sin \left(\frac{\pi x}{2}\right)+B, f^{\text{'}}\left(\frac{1}{2}\right)=\sqrt{2}$and${\int }_0^{1}f\left(x\right)dx=\frac{2A}{\mathrm{\pi }}$, then constants $A$ and $B$ are

The value of $\underset{0}{\overset{\pi /4}{\int }}\frac{\sin x+\cos x}{9+16\sin 2x}dx$ is

The value of $\underset{n\to \infty }{\lim }\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{6n}\right)$ is

$\underset{n\to \infty }{\lim }\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{6n}\right)$ is equal to