Question

Evaluate the following using limit as sum

$\int_{ - 1} ^{2} {e}^{2 - x} \: dx$
​

Answer 1

Answer:

Step-by-step explanation:

We know,

$\displaystyle\rm{\int^{b}_{a}\,f(x)\,dx=\lim_{n\to\infty}\dfrac{b-a}{n}\cdot\sum^{n}_{k=1}\,f\left\{a+\dfrac{(k-1)(b-a)}{n}\right\}}$

Now,

$\displaystyle\rm{\int^{2}_{-1}\,e^{\displaystyle2-x}\,dx=\lim_{n\to\infty}\dfrac{2-(-1)}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle2-\left\{-1+\dfrac{(k-1)(2-(-1))}{n}\right\}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle2-\left\{-1+\dfrac{3(k-1)}{n}\right\}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle2+1-\dfrac{3(k-1)}{n}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle3-\dfrac{3(k-1)}{n}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle3}\cdot\,e^{\displaystyle-\dfrac{3(k-1)}{n}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{\displaystyle3}}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle\dfrac{3(1-k)}{n}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{\displaystyle3}}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle\dfrac{3}{n}-\dfrac{3k}{n}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{\displaystyle3}}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle\dfrac{3}{n}}\cdot\,e^{\displaystyle-\dfrac{3k}{n}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{\displaystyle3}\cdot\,e^{\displaystyle\dfrac{3}{n}}}{n}\cdot\sum^{n}_{k=1}\,e^{\displaystyle-\dfrac{3k}{n}}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{3}\cdot\,e^{\frac{3}{n}}}{n}\cdot\sum^{n}_{k=1}\,\left(e^{-\frac{3}{n}}\right)^{k}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{3}\cdot\,e^{\frac{3}{n}}}{n}\cdot\left\{\left(e^{-\frac{3}{n}}\right)^{1}+\left(e^{-\frac{3}{n}}\right)^{2}+\left(e^{-\frac{3}{n}}\right)^{3}+\cdots+\left(e^{-\frac{3}{n}}\right)^{n}\right\}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{3}\cdot\,e^{\frac{3}{n}}}{n}\cdot\,\left(e^{-\frac{3}{n}}\right)^{1}\cdot\left\{1+e^{-\frac{3}{n}}+\left(e^{-\frac{3}{n}}\right)^{2}+\cdots+\left(e^{-\frac{3}{n}}\right)^{n-1}\right\}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{3}}{n}\cdot\,e^{\frac{3}{n}}\cdot\,e^{-\frac{3}{n}}\cdot\left\{1+e^{-\frac{3}{n}}+\left(e^{-\frac{3}{n}}\right)^{2}+\cdots+\left(e^{-\frac{3}{n}}\right)^{n-1}\right\}}$

$\displaystyle\rm{=\lim_{n\to\infty}\dfrac{3\,e^{3}}{n}\cdot\,1\cdot\dfrac{\left(e^{-\frac{3}{n}}\right)^{n}-1}{e^{-\frac{3}{n}}-1}}$

$\displaystyle\rm{=e^{3}\lim_{n\to\infty}\dfrac{3}{n}\cdot\dfrac{e^{-3}-1}{e^{-\frac{3}{n}}-1}}$

$\displaystyle\rm{=-e^{3}\left(e^{-3}-1\right)\cdot\lim_{n\to\infty}\dfrac{-\dfrac{3}{n}}{e^{-\frac{3}{n}}-1}}$

$\bf{We\,\,know\,,}\\\\\boxed{\displaystyle\lim_{x\to\infty}\dfrac{\dfrac{a}{x}}{e^{\frac{a}{x}}-1}=1}$

So,

$\displaystyle\rm{=-e^{3}\left(e^{-3}-1\right)\cdot1}$

$\displaystyle\rm{=-e^{3}\cdot\,e^{-3}+e^{3}\right)}$

$\displaystyle\rm{=e^{3}-1\right)}$