Integral as Limit of Sum

The value of $\underset{n\to \infty }{\lim }\sum _{r=1}^{r=4n}\frac{\sqrt{n}}{\sqrt{r}{\left(3\sqrt{r}+4\sqrt{n}\right)}^2}$ is equal to

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

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

Among

$\left(S1\right):\underset{n\to \infty }{\lim }\frac{1}{n^2}(2+4+6+\dots +2n)=1$

$\left(S2\right):\underset{n\to \infty }{\lim }\frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\dots +n^{15}\right)=\frac{1}{16}$

Consider:

Statement $1:$ $\underset{n\to \infty }{\lim }\frac{1}{n^2}\left(1+2+3+...+n\right)=1$

Statement $2:$$\underset{n\to \infty }{\lim }\frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+...+n^{15}\right)=\frac{1}{16}$

$\underset{n\to \infty }{\lim }{\left[\frac{\left({\displaystyle \frac{n}{3}+1}\right)\left({\displaystyle \frac{n}{3}}+2\right).....\left(n\right)}{{\left({\displaystyle \frac{n}{3}}\right)}^{{\displaystyle \raisebox{1ex}{$2n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\right]}^{\frac{3}{n}}$ is

$\underset{n\to \infty }{\lim }\frac{1}{2^n}\left(\frac{1}{\sqrt{1-{\displaystyle \frac{1}{2^n}}}}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{1-\frac{2}{2^n}}}}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{1-\frac{3}{2^n}}}}+...+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{1-\frac{2^n-1}{2^n}}}}\right)$ is equal to

Select the incorrect step while calculating definite integral ${\int }_1^3\left(x^2+e^{-x}\right)dx$ as limit of sums.

$\mathrm{I}:$$f\left(x\right)=x^2+e^{-x}, a=1, b=3, h=\frac{2}{n}$

$\mathrm{II}:$${\int }_1^3\left(x^2+e^{-x}\right)dx=2\underset{n\to \infty }{\lim }\frac{1}{n}\left[f\left(1\right)+f\left(1+\frac{2}{n}\right)+f\left(1+\frac{4}{n}\right)+...+f\left(1+\frac{2\left(n-1\right)}{n}\right)\right]$

$\mathrm{III}:$${\int }_1^3\left(x^2+e^{-x}\right)dx=2\underset{n\to \infty }{\lim }\frac{1}{n}\left[\left(1+e^{-1}\right)+\left\{{\left(1+\frac{2}{n}\right)}^2+e^{-\left(1+\frac{2}{n}\right)}\right\}+\left\{{\left(1+\frac{4}{n}\right)}^2+e^{-\left(1+\frac{4}{n}\right)}\right\}+...+\left\{{\left(1+\frac{2\left(n-1\right)}{n}\right)}^2+e^{-\left(1+\frac{2\left(n-1\right)}{n}\right)}\right\}\right]$

$\mathrm{IV}:$ ${\int }_1^3\left(x^2+e^{-x}\right)dx=2\underset{n\to \infty }{\lim }\frac{1}{n}\left[\left(1+{\left(1+\frac{2}{n}\right)}^2+{\left(1+\frac{4}{n}\right)}^2+...+{\left(1+\frac{2\left(n-1\right)}{n}\right)}^2\right)+\left\{e^{-1}+e^{-\left(1+\frac{2}{n}\right)}+e^{-\left(1+\frac{4}{n}\right)}+...+e^{-\left(1+\frac{2\left(n-1\right)}{n}\right)}\right\}\right]$

$\mathrm{V}:$ ${\int }_1^3\left(x^2+e^{-x}\right)dx=2\underset{n\to \infty }{\lim }\frac{1}{n}\left[\left(\frac{1-{\left(1+{\displaystyle \frac{2}{n}}\right)}^n}{1-\left(1+{\displaystyle \frac{2}{n}}\right)}\right)+\left\{\frac{e^{-1}\left(1-e^{-\left(1+{\displaystyle \frac{2}{n}}\right)n}\right)}{1-e^{-\left(1+{\displaystyle \frac{2}{n}}\right)}}\right\}\right]$

Evaluate ${\int }_2^3{x}^2 dx$ as the limit of a sum

$\underset{n\to \infty }{\lim }\left(\frac{1}{1+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\dots +\frac{n^4}{n^5+n^5}\right)=$

${\int }_0^1{a}^k{x}^{k}dx=$

For positive integer $n$, define $f\left(n\right)=n+\frac{16+5n-3n^2}{4n+3n^2}+\frac{32+n-3n^2}{8n+3n^2}+\frac{48-3n-3n^2}{12n+3n^2}+\dots +\frac{25n-7n^2}{7n^2}$. Then, the value of $\underset{n\to \infty }{\lim }f\left(n\right)$ is equal to

Let $f\left(x\right)=\underset{n\to +\infty }{\lim }{\left(\frac{n^n\left(x+n\right)\left(x+\frac{n}{2}\right)\cdots \left(x+\frac{n}{n}\right)}{n!\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right)\cdots \left(x^2+\frac{n^2}{n^2}\right)}\right)}^{\frac{x}{n}}$ for all $x>0$. Then

$\underset{n\to \infty }{\lim }\left[\frac{n^2}{\left(n^2+1\right)\left(n+1\right)}+\frac{{\displaystyle n^2}}{{\displaystyle \left(n^2+4\right)\left(n+2\right)}}+\frac{{\displaystyle n^2}}{{\displaystyle \left(n^2+9\right)\left(n+3\right)}}+...+\frac{{\displaystyle n^2}}{{\displaystyle \left(n^2+n^2\right)\left(n+n\right)}}\right]$ is equal to

The value of $\underset{n\to \infty }{\lim }\left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}............+\frac{n^2+n}{n-n^3}\right)$ is equal to 

The value of $\underset{n\to \infty }{\lim }\frac{1}{n^{n\lambda }}{\left[(n+1{)}^{\lambda }(n+2{)}^{\lambda }\dots (n+n{)}^{\lambda }\right]}^{\frac{1}{n}}$ is equal to

For $\alpha >-1$ and $\beta >-1$, the value of $\underset{n\to \infty }{\lim }\left(n^{\beta -\alpha }\right)\left(\frac{1^{\alpha }+2^{\alpha }+\cdots +n^{\alpha }}{1^{\beta }+2^{\beta }+\cdots +n^{\beta }}\right)$ is

Suppose $f$ is a differentiable function on $[0,1]$, such that the derivative of $f$ is continuous on $[0,1]$. Let $f\left(1\right)=6$ and $f\left(0\right)=1$ Then $\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^{n}f\left(\frac{k}{n}\right)f^{\text{'}}\left(\frac{k}{n}\right)$ is

The limit $\underset{n\to \infty }{\lim }\frac{1}{n^{2020}}\sum _{k=1}^n{k}^{2019}$