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#### Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 64 Maths Textbook Solution.

Answer:  $\frac{1}{2} \frac{e^{4}-1}{e^{2}}$

Hint: To solve the given statement we have to use the formula.

Given:  $\int_{-1}^{1} e^{2 x} d x$

Solution:

\begin{aligned} &\int_{a}^{b} f(x) d x=\frac{b-a}{n} \lim _{x \rightarrow \infty}[f(a)+f(a+h)+\ldots f(a+(h-1)] \\ & \end{aligned}

$f(x)=e^{2 x} \\$

$a=-1, b=1, h=\frac{1+1}{n}=\frac{2}{n}$

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

$=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2 a} e^{2 h}+\ldots .+e^{2 a} e^{2(n-1) h}\right] \\$

$=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}\left(1+e^{2 h}+\ldots .+e^{2(n-1) h}\right]\right. \\$

$=\frac{2}{n} e^{-2} \lim _{h \rightarrow \infty}\left[1+e^{2 h}+\ldots .+e^{2(n-1) h}\right]$

\begin{aligned} &a=1, r=e^{2 h} \\ & \end{aligned}

$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \\$

$\int_{1}^{1} e^{2 x} d x=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2 a+2 h}+\ldots .+e^{2 a+2(n-1) h}\right]$

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

$=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}\left(1+e^{2 h}+\ldots .+e^{2(n-1) h}\right]\right. \\$

$=\frac{2}{n e^{2}} \lim _{h \rightarrow \infty}\left[\frac{1 \cdot\left(e^{2 / n}-1\right)}{e^{2 h}-1}\right] \\$

$=\frac{2}{e^{2}} \lim _{h \rightarrow \infty}\left[\frac{1 \cdot\left(e^{2 m}-1\right)}{n e^{2 h}-1}\right]$

$=\frac{2}{e^{2}} \lim _{h \rightarrow \infty}\left[\frac{\left.e^{2 n} e^{2 / h}-1\right)}{\left(e^{2 n}-1\right) \frac{2}{h}}\right]$

\begin{aligned} &h=\frac{2}{n} \Rightarrow n=\frac{2}{h} \\ & \end{aligned}

$n \rightarrow \infty \\$

$n \rightarrow 0$

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

$=\frac{2}{n^{2}} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2 a} e^{2 h}+\ldots .+e^{2 a} e^{2(n-1) h}\right] \\$

$=\frac{2}{n^{2}} \lim _{h \rightarrow \infty}\left[e^{2 a}\left(1+e^{2 h}+\ldots .+e^{2(n-1) h}\right]\right. \\$

$=\frac{2}{n e^{2} \lim _{h \rightarrow \infty}}\left[\frac{1 \cdot\left(e^{2 m n}-1\right)}{e^{2 h}-1}\right] \\$

$=\frac{2}{e^{2}} \lim _{h \rightarrow \infty}\left[\frac{1 \cdot\left(e^{2 n n}-1\right)}{n e^{2 h}-1}\right]$

\begin{aligned} &=\frac{2}{e^{2}} \times\left(e^{4}-1\right) \lim _{n \rightarrow 0}\left[\frac{1}{\left(\frac{e^{2 n}-1}{h}\right) \times 2}\right] \\ & \end{aligned}

$=\frac{2}{e^{2}} \times\left(e^{4}-1\right) \lim _{n \rightarrow 0}\left[\frac{1}{\left(\frac{e^{2 n}-1}{h}\right) \times 4}\right]$

\begin{aligned} &=\frac{2\left(e^{4}-1\right)}{4 e^{2}} \lim _{n \rightarrow 0}\left[\frac{1}{\left(\frac{e^{2 n}-1}{h}\right)}\right] \\ & \end{aligned}

$=\frac{\left(e^{4}-1\right)}{2 e^{2}} \\$

$\int_{-1}^{1} e^{2 x} d x=\frac{1}{2} \frac{e^{4}-1}{e^{2}}$