Definite Integral as Limit of a Sum

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

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

Evaluate the following integral as a limit of sum :

${\int }_1^3{x}^{3}dx$

Evaluate the following integral as a limit of sum :

${\int }_0^3(e^x+6x^2)dx$

Evaluate the following integral as a limit of sum :

${\int }_0^1{3}^{x}dx$

Evaluate the following integral as a limit of sum :

${\int }_1^3{a}^{x}dx$

Evaluate the following integral as a limit of sum (up to two decimal places):

${\int }_0^4(x-x^2)dx$

Evaluate the following integral as a limit of sum :

${\int }_0^2(3x^2-1)dx$

Evaluate the following integral as a limit of sum :

${\int }_3^5(2-x)dx$

Evaluate the following integral as limit of a sum :

${\int }_0^2(2x+1)dx$

Suppose the limit $L=\underset{n\to \infty }{\lim }\sqrt{n}{\int }_0^1\frac{1}{{\left(1+x^2\right)}^n}dx$ exists and is larger than $\frac{1}{2}.$ Then,

$\underset{n\to \infty }{\lim }\frac{1+2^4+3^4+...............+n^4}{n^5}=$

$\underset{n\to \infty }{\lim }\left(\frac{\sqrt{1}+2\sqrt{2}+3\sqrt{3}+\dots +n\sqrt{n}}{n^{\frac{5}{2}}}\right)=$

Evaluate the following definite integral as a limit of sums

${\int }_0^2(x+4)dx$