Question

Evaluate the given definite integrals as limit of sums:
${\int }_0^4(x+e^{2x})dx$

Answer 1

${\int }_a^{b}f\left(x\right)dx=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(a)+f(a+h)+...+f(a+(n-1)h\left)\right]$, where $h=\frac{b-a}{n}$
Here, $a=0,b=4$, and $f\left(x\right)=x+e^{2x}$
$\therefore h=\frac{4-0}{n}=\frac{4}{n}$
$\Rightarrow {\int }_0^4(x+e^{2x)}dx=(4-0)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(0)+f(h)+f(2h)+...+f((n-1)h\left)\right]$
$=4\underset{n\to \infty }{lim}\frac{1}{n}\left[\right(0+e^0)+(h+e^{2h})+(2h+e^{2\cdot 2h})+....+\left\{\right(m-1)h+e^{2(n-1)h}\}]$
$=4\underset{n\to \infty }{lim}\frac{1}{n}[1+(h+e^{2h})+(2h+e^{4h})+...+\left\{\right(m-1)h+e^{2(n-1)h}\}]$
$=4\underset{n\to \infty }{lim}\frac{1}{n}[\{h+2h+3h+....+(n-1\left)h\right\}+(1+e^{2h}+e^{4h}+...+e^{2(n-1)h}\left)\right]$
$=4\underset{n\to \infty }{lim}\frac{1}{n}[h\{1+2+...(n-1\left)\right\}+\left(\frac{2^{2hn}-1}{e^{2h}-1}\right)]$
$=4\underset{n\to \infty }{lim}\frac{1}{n}[\frac{h(n-1)n)}{2}+\left(\frac{e^{2hn}-1}{e^{2h}-1}\right)]$
$=\underset{n\to \infty }{lim}\frac{1}{n}[\frac{4}{n}\cdot \frac{(n-1)n}{2}+\left(\frac{e^8-1}{e^{\frac{8}{n}}-1}\right)]$
$=4\left(2\right)+4\underset{n\to \infty }{lim}\frac{(e^8-1}{\left(\frac{e^{\frac{8}{n}}-1}{\frac{8}{n}}\right)8}$
$=8+\frac{4\cdot (e^8-1)}{8},(\because \underset{n\to \infty }{lim}\frac{e^x-1}{x}=1)$
$=8+\frac{e^8-1}{2}=\frac{15+e^8}{2}$