Evaluate the following definite integral as limit of sum: ^ b _ a… - Vidyadip

Question

Evaluate the following definite integral as limit of sum:
$\int\limits^{\text{b}}_{\text{a}}\text{x dx}$

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Answer

We know that,$\int\limits^{b}_\text{a}\text{f}\big(\text{x}\big) \text{dx}=\lim\limits_{\text{h} \rightarrow 0}\text{h}\big[\text{f}(\text{a)}+\text{a}(\text{a+h})+\text{f}(\text{a+2h})+.....+\text{f}(\text{a}+\text+{(\text{n}-1)\text{h})}\big]\text{where nh = b - a}$Here, $\text{a = a, b = b}\text{ and } \text{f}(\text{x})=\text{x}$
$\therefore \int_\limits\text{a}^{b} \text{x} \text { dx}=\lim_\limits{\text{h}\rightarrow0} \text{h} \big[\text{a+(a+h)+(a+2h)+......+(a+(n-1)h}\big]$
$\Rightarrow \int_\limits{a}^{b} \text{x} \text{ dx}=\lim_\limits{\text{h}\rightarrow0} \text{h} \big[\text{na}+\text{h}(1+2+3+..........+(\text{n}-1))\big]$
$\Rightarrow \int_\limits{a}^{b}\text{x}\text{ dx}\lim\limits_{\text{h}\rightarrow0}\bigg[\text{anh}+\text{h}\frac{\text{n}\text{(n}-1)}{2}\bigg]=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{anh}+\frac{\text{nh}{\text{(nh-h)}}}{2}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{a}(\text{b}-\text{a})+\frac{(\text{b}-\text{a})(\text{b}-\text{a}-\text{h})}{2}\bigg][\because\text{nh}=\text{b}-\text{a}]$ $=\bigg[\text{a}\text{(b}-\text{a)}+\frac{\text{(b}-\text{a)}\text{(b}-\text{a)}}{2}\bigg]=\text{(b}-\text{a)}\bigg[\text{a}+\frac{\text{(b}-\text{a)}}{2}\bigg]=\text{(b}-\text{a)}\bigg[\frac{2\text{a}+\text{b}-\text{a}}{2}\bigg]$$=\frac{\text{(b}-\text{a)}\text{(b}+\text{a)}}{2}=\frac{\text{b}^{2}-\text{a}^{2}}{2} $

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