#### Explain solution rd sharma class 12 chapter 19 Definite Integrals exercise 19.5, question 15

$e^{2}-1$

Hint:

To solve the given statement, we have to use the formula of addition limits.

Given:

$\int_{2}^{0}(e^{x})dx$

Solution:

We have,

$\int_{a}^{b}\left ( fx \right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$

Where, $h=\frac{b-a}{n}$

Here,

$a=0,b=2,f(x)=e^{x}\\ h=\frac{2-0}{n}=\frac{2}{n},nh=2$

Thus, we have

\begin{aligned} I &=\int_{0}^{2}\left(e^{x}\right) d x \\ I &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f((n-1)) h] \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} h\left[1+e^{h}+e^{2 h}+\ldots+e^{(n-1) h}\right] \\ &=\lim _{h \rightarrow 0} h\left[\frac{\left(e^{h}\right)^{n}-1}{e^{h}-1}\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow 0} h\left[\frac{e^{n h}-1}{e^{h}-1}\right] \\ &=\lim _{n \rightarrow 0} h\left[\frac{e^{2}-1}{e^{h}-1}\right] \quad[n h=2] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow 0}\left[\frac{e^{2}-1}{e^{h}-1}{h}\right] \\ &=e^{2}-1 &\left[\mathrm{Q} \frac{e^{a}-1}{a} \Rightarrow 1\right] \end{aligned}