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# Show that

Question:

Evaluate $\int_{0}^{1} e^{2-3 x} d x$ as a limit of a sum.

Solution:

Let $I=\int_{0}^{1} e^{2-3 x} d x$

It is known that,

$\int_{0}^{b} f(x) d x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[f(a)+f(a+h)+\ldots+f(a+(n-1) h)]$

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

Here, $a=0, b=1$, and $f(x)=e^{2-3 x}$

$\Rightarrow h=\frac{1-0}{n}=\frac{1}{n}$

$\therefore \int_{0}^{1} e^{2-3 x} d x=(1-0) \lim _{n \rightarrow \infty} \frac{1}{n}[f(0)+f(0+h)+\ldots+f(0+(n-1) h)]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}+e^{2-3 h}+\ldots e^{2-3(n-1) n}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}\left\{1+e^{-3 h}+e^{-6 h}+e^{-9 h}+\ldots e^{-3(n-1) h}\right\}\right]$

$=\lim _{h \rightarrow \infty} \frac{1}{n}\left[e^{2}\left\{\frac{1-\left(e^{-3 h}\right)^{n}}{1-\left(e^{-3 h}\right)}\right\}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}\left\{\frac{1-e^{-\frac{3}{n} \times n}}{1-e^{-\frac{3}{n}}}\right\}\right]$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{e^{2}\left(1-e^{-3}\right)}{1-e^{-\frac{3}{n}}}\right]$

$=e^{2}\left(e^{-3}-1\right) \lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{1}{e^{-\frac{3}{n}}-1}\right]$

$=e^{2}\left(e^{-3}-1\right) \lim _{n \rightarrow \infty}\left(-\frac{1}{3}\right)\left[\frac{-\frac{3}{n}}{e^{-\frac{3}{n}}-1}\right]$

$=\frac{-e^{2}\left(e^{-3}-1\right)}{3} \lim _{n \rightarrow \infty}\left[\frac{-\frac{3}{n}}{e^{-\frac{3}{n}}-1}\right]$

$=\frac{-e^{2}\left(e^{-3}-1\right)}{3}(1)$                 $\left[\lim _{n \rightarrow \infty} \frac{x}{e^{x}-1}\right]$

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

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