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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate: $\int\limits_1^3(\text{x}^2+3\text{x}+\text{e}^\text{x})\text{dx},$ as the limit of the sum.
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Answer

$\int\limits_1^3(\text{x}^2+3\text{x}+\text{e}^\text{x})\text{dx}$
$\lim\limits_{\text{h} \rightarrow 0}\ \text{h}\big[\text{f}(1)+\text{f}(1+\text{h})+\text{f}(1+2\text{h})+\ ....\ +\text{f}(1+2(\text{n}-1)\text{h})\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[\big(1+3+\text{e})+((1+\text{h})^2+3(1+\text{h})+\text{e}^{1+\text{h}}\big)\\+\big((1+2\text{h})^2+3(1+2\text{h})+\text{e}^{1+2\text{h}}\big)+\ ...\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[4+\text{e}+\big(1+\text{h}^2+2\text{h}+3+3\text{h}+\text{e}^{1+\text{h}}\big)\\+\big(1+4\text{h}^2+4\text{h}+3+6\text{h}+\text{e}^{1+2\text{h}}\big)+\ ...\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[4+\text{e}+\big(4+\text{h}^2+5\text{h}+\text{e}^{1+\text{h}}\big)\\+\big(4+4\text{h}^2+10\text{h}+\text{e}^{1+2\text{h}}\big)+\ ...\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[4\text{n}+\text{e}\big(1+\text{e}^\text{h}+\text{e}^{2\text{h}}+\ ...\big)\\+\text{h}^2\big[1^2+2^2+ ...\big]+5\text{h}(1+2+\ ...)\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\Bigg[4\text{n}+\text{e}\bigg(\frac{1(\text{e}^\text{hn}-1)}{\text{e}^\text{h}-1}\bigg)+\text{h}^2\bigg(\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg)+5\text{h}\bigg(\frac{\text{n}(\text{n}-1)}{2}\bigg)\Bigg]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\Bigg[4\text{n}+\text{e}\bigg(\frac{\text{e}^\text{nh}-1}{\text{e}^\text{h}-1}\bigg)+\frac{\text{h}^2\text{n}^3}{6}\bigg(1-\frac{1}{\text{n}}\bigg)\bigg(2-\frac{1}{\text{n}}\bigg)+\frac{5\text{hn}^2}{2}\bigg(1-\frac{1}{\text{n}}\bigg)\Bigg]$
$\lim\limits_{\text{n}\rightarrow\infty}\ \frac{2}{\text{n}}\Bigg[4\text{n }+\text{e}\Bigg(\frac{\text{e}^{\frac{\text{n}\times2}{\text{n}}}}{\text{e}^\frac{2}{\text{n}}-1}\Bigg)+\frac{4}{\text{n}^2}\frac{\text{n}^3}{6}\bigg(1-\frac{1}{\text{n}}\bigg)\bigg(2-\frac{1}{\text{n}}\bigg)+\frac{5}{2}\times\text{n}^2\times\frac{2}{\text{n}}\bigg(1-\frac{1}{\text{n}}\bigg)\Bigg]$
$\lim\limits_{\text{n}\rightarrow\infty}\ \frac{2}{\text{n}}\Bigg[4\text{n}+\text{e}\bigg(\frac{\text{e}^2-1}{\text{e}^\frac{2}{\text{n}}-1}\bigg)+\frac{4\text{n}}{6}\bigg(1-\frac{1}{\text{n}}\bigg)\bigg(2-\frac{1}{\text{n}}\bigg)+5\text{n}\bigg(1-\frac{1}{\text{n}}\bigg)\Bigg]$
$\lim\limits_{\text{n}\rightarrow\infty}\ 2\Bigg[4+\frac{\text{e}}{\text{n}}\frac{(\text{e}^2-1)}{\text{e}^\frac{2}{\text{n}}-1}\bigg)+\frac{2}{3}\bigg(1-\frac{1}{\text{n}}\bigg)\bigg(2-\frac{1}{\text{n}}\bigg)+5\bigg(1-\frac{1}{\text{n}}\bigg)\Bigg]$
$=8+\text{e}(\text{e}^2-1)+\frac{4}{3}+5$
$=\frac{24+4+15}{3}+\text{e}(\text{e}^2-1)$
$=\frac{43}{3}+\text{e}(\text{e}^2-1)$

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