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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals as limit of sum:$\int\limits^{5}_{0}(\text{x}+1)\text{dx}$
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Answer

$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=5,\text{ f(x)}=\text{x}+1,\text{ h}=\frac{5-0}{\text{n}}=\frac{5}{\text{n}}$
Therefore, $\text{I}=\int\limits^{5}_{0}(\text{x}+1)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+1)+(\text{h}+1)+\ ....+\ \big\{(\text{n}-1)\text{h}+1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{5}{\text{n}}\Big[\text{n}+\frac{5(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}5\Big[1+\frac{5}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=5+\frac{25}{2}$
$=\frac{35}{2}$

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