Question

State whether the given statement is True or False

${\int }_0^2{e}^{x^2}dx$ can be represented as $2\underset{n\to \infty }{lim}\frac{1}{n}[e^0+e^{\frac{4}{n^2}}+e^{\frac{16}{n^2}}+......+e^{\frac{2(n-1{)}^2}{n^2}}]$

• A. True
• B. False

Answer 1

The correct option is A True

We have to state whether ${\int }_0^2{e}^{x^2}dx$ can be represented as $2{lim}_{n\to 0}\frac{1}{n}[e^0+e^{\frac{4}{n^2}}+e^{\frac{16}{n^2}}+...+e^{\frac{2(n-1{)}^2}{n^2}}]$ is true or false.
We know that ${\int }_a^{b}f\left(x\right)dx={lim}_{n\to \infty }\left[\frac{b-a}{n}{\sum }_{k=1}^{n}f(a+k\frac{b-a}{n})\right]$
Here we have $f\left(x\right)=e^{x^2},a=0,b=2$
Therefore ${\int }_0^2{e}^{x^2}dx={lim}_{n\to \infty }\left[\frac{2-0}{n}{\sum }_{k=1}^{n}f(0+k\frac{2-0}{n})\right]$
$={lim}_{n\to \infty }\left[\frac{2}{n}{\sum }_{k=1}^{n}f\left(\frac{2k}{n}\right)\right]$
$=2{lim}_{n\to \infty }\left[\frac{1}{n}{\sum }_{k=0}^{n-1}f\left(\frac{2k}{n}\right)\right]$
$=2{lim}_{n\to \infty }\frac{1}{n}\left[f\right(0)+f\left(\frac{2}{n}\right)+f\left(\frac{4}{n}\right)+...+f\left(\frac{2(n-1)}{n}\right)]$
Since $f\left(x\right)=e^{x^2}$ we get
${\int }_0^2{e}^{x^2}dx=2{lim}_{n\to \infty }\frac{1}{n}[e^0+e^{\frac{4}{n^2}}+e^{\frac{16}{n^2}}+...+e^{\frac{2(n-1{)}^2}{n^2}}]$
Hence the answer is TRUE.