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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS3 Marks
Question
Evaluate the following as limit of sums:
$\int\limits^2_0\text{e}^\text{x}\text{dx}$
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Answer

We know that $\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\lim\limits_{\text{r}=0}\text{h}\sum\limits^{\text{n}-1}_{\text{r}=0}\text{f}(\text{a}+\text{rh})$
For $\text{I}=\int\limits^2_0\text{e}^\text{x}\text{dx},$ we have a = 0 and b = 2
$\therefore\ \text{h}=\frac{\text{b}-\text{a}}{\text{n}}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
$\therefore\ \text{I}=\int\limits^2_0\text{e}^\text{x}\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[1+\text{e}^\text{h}+\text{e}^{2\text{h}}+\dots+\text{e}^{(\text{n}-1)^\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\frac{1\cdot(\text{e}^\text{h})^\text{n}-1}{\text{e}^\text{h}-1}\Big]=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big(\frac{\text{e}^{\text{eh}}-1}{\text{e}^\text{h}-1}\Big)$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big(\frac{\text{e}^2-1}{\text{e}^\text{h}-1}\Big)$
$=\text{e}^2=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{e}^\text{h}-1}=\text{e}^2-1$

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