Evaluate the following definite integral as limit of sum: _0^ 4 (… - Vidyadip

Question

Evaluate the following definite integral as limit of sum:$\int_\limits{0}^{4}\text{(x}+\text{e}^{2\text{x}})\ \text{dx}$

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Answer

$\text{we}\ \text{know}\ \text{that}\ \int\limits_{\text{a}}^{\text{b}}\text{f}\text{(x)}\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[\text{f}\ \text{(a)}+\text{f}\text{(a+h)}+\text{f}\text{(a}+2\text{h})+.....+\text{f}\text{(a}+\text{(n}-1)\text{h)}\big] $ $\text{where}\ \text{nh}=\text{b}-\text{a}$ $\text{Here},\ \text{a}=0,\text{b}=4,\text{nh}=4\ \text{and}\ \text{f}\ \text{(x)}=\text{x}+\text{e}^{2\text{x}}$ $\therefore\ \ \int\limits_{0}^{4}\text{(x+e}^{2\text{h}})\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\bigg[1+\text{(h}+\text{e}^{2\text{h}})+\text{(2h}+\text{e}^{4\text{h}})+....+\big(\text{(n}-1)\text{h+e}^{2\text{(n-1)}\text{h}}\big)\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{(h}+2\text{h}+.....+\text{(n}-1)\text{h)}+\big(1+\text{e}^{2\text{h}}+\text{e}^{4\text{h}}+....+\text{e}^{2(\text{n}-1)\text{h}}\big)\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}(1+2+......+\text{(n}-1)+\text{a}\bigg(\frac{\text{r}^{\text{n}}-1}{\text{r}-1}\bigg)\bigg]=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{h.h}\frac{\text{n}\text{(n}-1)}{2}+\frac{1\big(\text{e}^{2\text{h}^{\text{n}}}\big)-1}{\text{e}^{2\text{h}}-1}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}=\bigg[\frac{\text{nh}\text{(nh}-\text{h)}}{2}+\frac{\text{h}\big(\big(\text{e}^{2\text{nh}}\big)-1\big)}{\text{e}^{2\text{h}}-1}\bigg] =\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{4(4-\text{h)}}{2}+\frac{\text{h}\big(\big(\text{e}^{2.4}\big)-1\big)}{\text{e}^{2\text{h}}-1}\bigg]$$=\frac{4(4-0)}{2}+\big(\text{e}^{8}-1\big)\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{e}^{2\text{h}}-1}=8\big(\text{e}^{8}-1\big)\frac{1}{2}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{e}^{2\text{h}}-1}=8+\frac{\big(\text{e}^{8}-1\big)}{2}$
$=\bigg[\because\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{e}^{\text{x}}-1}=1\bigg]$

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