Question

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

Answer 1

We know that ${\int }_a^{b}f\left(x\right)dx={lim}_{h\to 0}h\left[f\right(a)+f(a+h)+f(a+2h)+.....+f\{a+(n-1\left)h\right\}]$
where, nh =b-a
Given,${\int }_0^4(x+e^{2x})dx$
Here, a=0, b=4 and nh =4
and $f\left(x\right)=(x+e^{2x}),\therefore f\left(0\right)=(0+e^{2\left(0\right)})=1,$
$f(0+h)=h+e^{2h},f(0+2h)=2h+e^{4h},$
and $f(0+(n-1\left)h\right)=(n-1)h+e^{2(n-1)h}$
$\therefore {\int }_0^4(x+e^{2x})dx={lim}_{h\to 0}h[1+(h+e^{2h})+(2h+e^{4h})+...+(n-1)h+e^{2(n-1)h}]={lim}_{h\to 0}h[\{h+2h+3h+...+(n-1\left)\right\}h+\{1+e^{2h}+e^{4h}+...+e^{2(n-1)h}\}]$
$={lim}_{h\to 0}h[h\{1+2+3+...+(n-1\left)\right\}+\{1+e^{2h}+e^{4h}+...+e^{2nh-2h}\}]={lim}_{h\to 0}h\{h\frac{(n-1)n}{2}+1\left(\frac{(e^{2h}{)}^n-1}{e^{2h}-1}\right)\}(\because \sum n=\frac{n(n+1)}{2})$
$(\because 1+e^{2h}+e^{4h}+...+e^{2(n-1)h})$ is a GP series whose first term is 1 and common ratio $r=\frac{e^{2h}}{1}=e^{2h}$
$\therefore$ Sum $=\frac{a(r^n-1)}{r-1}$

$={lim}_{h\to 0}\{h^2\frac{(n-1)n}{2}+h\frac{(e^{2nh}-1)}{e^{2h}-1}\}={lim}_{h\to 0}\frac{(nh-h)\left(nh\right)}{2}+{lim}_{h\to 0}\frac{(e^2{)}^4-1}{\frac{e^{2h}-1}{2h}\times 2}={lim}_{h\to 0}\{\frac{(4-h)4}{2}+\frac{e^8-1}{2}\}(\because {lim}_{h\to 0}\frac{e^x-1}{x}=1)=\frac{4\times 4}{2}+\frac{e^8-1}{2}=(8-\frac{1}{2})+\frac{e^8}{2}=\frac{15+e^8}{2}$