Question

Evaluate ${\int }_0^1{e}^{2-3x}dx$ as a limit of a sum.

Answer 1

${\int }_a^{b}f\left(x\right)dx={lim}_{h\to 0}h[f\left(a\right)+f(a+h)+f(a+2h)+.....+f(a+(n-1\left)\right)h]$

Where, $h=\frac{b-a}{n}$

$\therefore nh=1-0=1.$

$\Rightarrow I={\int }_0^1{e}^{2-3x}dx$

$={lim}_{h\to 0}h[f\left(0\right)+f\left(h\right)+f\left(2h\right)+.....+f(n-1)h]$

$={lim}_{h\to 0}h[e^2+e^{2-3h}+e^{2-3\left(2h\right)}+.....+e^{2-3(n-1)h}]$

$={lim}_{h\to 0}h{e}^2[1+e^{-3h}+e^{-3\left(2h\right)}+.....+e^{-3(n-1)h}]$

$={lim}_{h\to 0}h{e}^2\left\{\frac{{{(e}^{-3h})}^n-1}{e^{-3h}-1}\right\}$

$={lim}_{h\to 0}h{e}^2\left\{\frac{e^{-3nh}-1}{e^{-3h}-1}\right\}$

$=e^2{lim}_{h\to 0}\frac{e^{-3}-1}{\left(\frac{e^{-3h}-1}{-3h}\right)}\times \frac{-1}{3}$

$\Rightarrow I=e^2(e^{-3}-1)\times \frac{-1}{3}$

$=\frac{1}{3}(e^2-e^{-1})$

$[\because {lim}_{h\to 0}\frac{e^{-3h}-1}{-3h}=1]$ $[\because nh=1].$

Hence, solved.