Evaluate the following integrals as limit of sum: ^1_ 0 (3x^2+5x … - Vidyadip

Question

Evaluate the following integrals as limit of sum:
$\int\limits^1_{0}\big(3\text{x}^2+5\text{x}\big)\text{dx}$

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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, $\text{a}=0,\text{ b}=1,\text{ f(x)}=3\text{x}^2+5\text{x},\text{ h}=\frac{1-0}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^1_{0}\big(3\text{x}^2+5\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+0)+(3\text{h}^2+5\text{h})+\ \\ .....\ +\big\{3(\text{n}-1)^2\text{h}^2+5(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[5\text{h}(1+2+\ ....\ +\text{n})\\+3\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[5\text{h}\frac{\text{n}(\text{n}-1)}{2}+\text{h}^2\frac{3\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\frac{5(\text{n}-1)}{2}+\frac{(\text{n}-1)(2\text{n}-1)}{2\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg[\frac{5}{2}\Big(1-\frac{1}{\text{n}}\Big)+\frac{1}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg]$
$=\frac{5}{2}+1$
$=\frac{7}{2}$

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