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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate $\int^{1}_{0}\text{e}^{2-3\text{x}}\text{dx}$ as a limit of a sum:
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Answer

$\text{Let I}=\int^{1}\limits_{0}\text{e}^{2-3\text{x}}\text{dx}$
It is know that,
$\int^{1}\limits_{0}\text{f (x)dx}=(\text{b}-\text{a})\lim\limits_{\text{n}_\rightarrow\infty}\frac{1}{\text{n}}\Big[\text{f(a)}+\text{f(a}+\text{h)}+...+\text{f (a}+(\text{n}-1)\text{h})\Big]$
Where,$\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
H,ere $\text{a}=0,\text{b}=1, $ and $\text{f (x)}=\text{e}^{2-3\text{x}}$
$\Rightarrow\text{h}=\frac{1-0}{\text{n}}=\frac{1}{\text{n}}$
$\therefore\int^{1}\limits_{0}\text{e}^{2-3\text{x}}\text{dx}=(1-0)\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\Big[\text{f}(0)+\text{f}(0+\text{h})+...+\text{f}(0+(\text{n}-1)\text{h)}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\Big[\text{e}^{2}+\text{e}^{2-3\text{h}}+...\text{e}^{2-3(\text{n}-1)\text{h}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\Big[\text{e}^{2}\left\{1+\text{e}^{-3\text{h}}+...\text{e}^{-6\text{h}}+\text{e}^{-9\text{h}}+...\text{e}^{-3(\text{n}-1)\text{h}}\right\}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\Bigg[\text{e}^{2}\left\{\frac{1-(\text{e}^{-3\text{h)}}{\text{h}}}{1-\text{(e}^{-3\text{h}})}\right\}\Bigg]$
$=\lim_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\begin{bmatrix}\text{e}^{2}\left\{\frac{1-\text{e}^{-\frac{3}{\text{n}}\times\text{n}}}{1-\text{e}^{-\frac{3}{\text{n}}}}\right\} \end{bmatrix} $
$=\lim_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\begin{bmatrix}\frac{\text{e}^{2}(1-\text{e}^{-3})}{1-\text{e}^{-\frac{3}{\text{n}}}}\end{bmatrix} $
$\text{e}^{2}(\text{e}^{-3}-1)=\lim_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}=\Bigg[\frac{1}{\text{e}^{-\frac{3}{\text{n}}}-1}\Bigg] $
$\text{e}^{2}(\text{e}^{-3}-1)=\lim_{\text{n}\rightarrow\infty}\bigg(-\frac{1}{3}\bigg)\Bigg[\frac{-\frac{3}{\text{n}}}{\text{e}^{-\frac{3}{\text{n}}}-1}\Bigg] $
$=\frac{\text{e}^{2}(\text{e}^{-3}-1)}{3}=\lim_{\text{n}\rightarrow\infty}\Bigg[\frac{-\frac{3}{\text{n}}}{\text{e}^{-\frac{3}{\text{n}}}-1}\Bigg] $
$=\frac{\text{e}^{2}(\text{e}^{-3}-1)}{3}\ \ \ \ \ \ \Bigg[\lim_{\text{n}\rightarrow\infty}\frac{\text{x}}{\text{e}^{\text{x}}-1}\Bigg] $
$=\frac{-\text{e}^{-1}+\text{e}^{2}}{3}$
$=\frac{1}{3}\bigg(\text{e}^{2}-\frac{1}{\text{e}}\bigg)$

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