Question

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

Answer 1

Let $I={\int }_0^1{e}^{2-3x}dx$
It is known that,
${\int }_a^{b}f\left(x\right)dx=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(a)+f(a+h)+...+f(a+(n-1)h]$
Where, $h=\frac{b-a}{n}$
Here, $a=0,b=1$, and $f\left(x\right)=e^{2-3x}$
$\Rightarrow h=\frac{1-0}{n}=\frac{1}{n}$
$\therefore {\int }_0^1{e}^{2-3x}dx=(1-0)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(0)+f(0+h)+...+f(0+(n-1)h\left)\right]$
$=\underset{n\to \infty }{lim}\frac{1}{n}[e^2+e^{2-3h}+....e^{2-3(n-1)h}]$
$=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\{1+e^{-3h}+e^{-6h}+e^{-9h}+...e^{-3(n-1)h}\}\right]$
$=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-(e^{-3h}{)}^n}{1-\left(e^{-3h}\right)}\right\}\right]$
$=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-e^{-\frac{3}{n}\times n}}{1-e^{-\frac{3}{n}}}\right\}\right]$
$=\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2(1-e^{-3})}{1-e^{-\frac{3}{4}}}\right]$
$=e^2(e^{-3}-1)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{-\frac{3}{n}}-1}\right]$
$=e^2(e^{-3}-1)\underset{n\to \infty }{lim}(-\frac{1}{3})\left[\frac{-\frac{3}{n}}{e^{-\frac{3}{n}}-1}\right]$
$=\frac{-e^2(e^{-3}-1)}{3}\underset{n\to \infty }{lim}\left[\frac{-\frac{3}{n}}{e^{-\frac{3}{n}}-1}\right]$
$=\frac{-e^2(e^{-3}-1)}{3}\left(1\right)\left[\underset{n\to \infty }{lim}\frac{x}{e^x-1}\right]$
$=\frac{-e^{-1}+e^2}{3}$
$=\frac{1}{3}(e^2-\frac{1}{e})$