Question

Express ${\int }_0^2{e}^{x}dx$ as the limit of a sum.

Answer 1

Consider the problem

We know that

$\int f\left(x\right)dx=\left(a-b\right)\underset{n\to \infty }{lim}\frac{1}{n}\left(f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)...+f\left(a+\left(n-1\right)h\right)\right)$

So, Putting

$a=0$

$b=2$

$\begin{array}{c}h=\frac{b-a}{n}\\ =\frac{2-0}{n}\\ =\frac{2}{n}\end{array}$

$f\left(x\right)=e^x$

Hence, we can write,

${\int }_0^{2}exdx=\left(2-0\right)\underset{n\to \infty }{lim}\frac{1}{n}\left(f\left(0\right)+f\left(0+h\right)+f\left(0+2h\right)\cdots \dots +f\left(0+\left(n-1\right)h\right)\right)$

$\begin{array}{c}f\left(x\right)=e^x\\ f\left(0\right)=e^0=1\\ f\left(0+h\right)=e^{0+h}=e^h\\ f\left(0+2h\right)=e^{0+2h}=e^{2h}\\ f\left(0+\left(n-1\right)h\right)=e^{0+\left(n-1\right)h}=e^{\left(n-1\right)h}\end{array}$

Thus, equation becomes

$\begin{array}{c}\therefore {\int }_0^{2}exdx=\left(2-0\right){lim}_{n\to \infty }\frac{1}{n}\left(f\left(0\right)+f\left(0+h\right)+f\left(0+2h\right)+\cdots \cdots +f\left(0+\left(n-1\right)h\right)\right)\\ =2.{lim}_{n\to \infty }\frac{1}{n}\left(1+e^h+e^{2h}+......+e^{\left(n-1\right)h}\right)\end{array}$

Let,

$s=e^h+e^{2h}+\dots \dots +e^{\left(n-1\right)h}$

it is a $G.P$ with common ratio $\left(r\right)$

Sum of $G.P$

$s=\frac{a\left(1-r^n\right)}{1-r}=\frac{e^h\left(1-{\left(e^h\right)}^n\right)}{1-e^h}=\frac{e^h\left(1-e^{nh}\right)}{1-e^h}$

Thus,

$\therefore {\int }_0^2{e}^{x}dx=2.\underset{n\to \infty }{lim}\frac{1}{n}\left(1+e^h+e^{2h}+\dots \dots +e^{\left(n-1\right)h}\right)$

Now, putting the value of $s$

$\begin{array}{c}{\int }_0^2{e}^{x}dx=2.{lim}_{n\to \infty }\frac{1}{n}\left(1+\frac{e^h\left(1-e^{nh}\right)}{1-e^h}\right)\\ =2\left({lim}_{n\to \infty }\frac{1}{n}+{lim}_{n\to \infty }\frac{1}{n}\left(e^h\left(\frac{1-e^{nh}}{1-e^h}\right)\right)\right)\\ =2\left(\frac{1}{\infty }+{lim}_{n\to \infty }\left(\frac{e^h}{n}.\frac{e^{nh}-1}{e^h-1}\right)\right)\left(taking-1commonnumeratoranddenominator\right)\\ =2\left(0+{lim}_{n\to \infty }\left(\frac{e^h}{n}.\frac{e^{nh}-1}{h.\frac{e^h-1}{h}}\right)\right)\left(Multiplyinganddividingdenominatorbyh\right)\\ =2.{lim}_{n\to \infty }\frac{e^h}{n}\left(\frac{e^{nh}-1}{h}\right)\left(\frac{1}{\frac{e^h-1}{h}}\right)\end{array}$

$=2.\underset{n\to \infty }{lim}\frac{e^h}{n}\left(\frac{e^{nh}-1}{h}\right).\underset{n\to \infty }{lim}\left(\frac{1}{\frac{e^h-1}{h}}\right)$

Now, solving $\underset{n\to \infty }{lim}\frac{1}{\frac{e^h-1}{h}}$

As

$\begin{array}{c}n\to \infty \\ \Rightarrow \frac{2}{h}\to \infty \\ \Rightarrow h\to 0\end{array}$

Therefore,

$\begin{array}{c}{lim}_{n\to \infty }\frac{1}{\frac{e^h-1}{h}}={lim}_{h\to 0}\frac{1}{\frac{e^h-1}{h}}\\ =\frac{1}{1}=1\end{array}$

Thus, the equation becomes

$\begin{array}{c}\therefore {\int }_0^2{e}^{x}dx=2{lim}_{n\to \infty }\frac{e^h}{n}\left(\frac{e^{nh}-1}{h}\right){lim}_{n\to \infty }\left(\frac{1}{\frac{e^h-1}{h}}\right)\\ =2{lim}_{n\to \infty }\frac{e^h}{n}\left(\frac{e^{nh}-1}{h}\right)1\\ =2{lim}_{n\to \infty }\frac{e^{\frac{2}{n}}}{n}\left(\frac{e^{n.\frac{2}{n}}-1}{\frac{2}{n}}\right)\left(u\sin gh=\frac{2}{n}\right)\\ =2{lim}_{n\to \infty }{e}^{\frac{2}{n}}\left(\frac{e^2-1}{2}\right)\\ =2\left(e^{\frac{2}{\infty }}\left(\frac{e^2-1}{2}\right)\right)\\ =2e^0\frac{\left(e^2-1\right)}{2}\\ =\frac{2}{2}1\left(e^2-1\right)\\ =e^2-1\end{array}$