Question Number 5406 by 1771727373 last updated on 14/May/16
$$\int_{−\infty} ^{\infty} \:{e}^{{x}} {dx}=? \\ $$
Commented by FilupSmith last updated on 14/May/16
![S=lim_(k→∞) ∫_(−k) ^k e^x dx ∫e^x dx=e^x +c S=lim_(k→∞) [e^x +c]_(−k) ^k S=lim_(k→∞) ((e^k +c)−(e^(−k) +c)) S=lim_(k→∞) (e^k −e^(−k) ) S=lim_(k→∞) (e^k −(1/e^k )) S=lim_(k→∞) ((e^(2k) −1)/e^k ) use L′Hopital′s Law S=lim_(k→∞) (e^(2k) /(2e^k )) S=lim_(k→∞) (e^k /2) ∴ S=∞](https://www.tinkutara.com/question/Q5408.png)
$${S}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\:\underset{−{k}} {\overset{{k}} {\int}}{e}^{{x}} {dx} \\ $$$$\int{e}^{{x}} {dx}={e}^{{x}} +{c} \\ $$$$ \\ $$$${S}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left[{e}^{{x}} +{c}\right]_{−{k}} ^{{k}} \\ $$$${S}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left(\left({e}^{{k}} +{c}\right)−\left({e}^{−{k}} +{c}\right)\right) \\ $$$${S}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left({e}^{{k}} −{e}^{−{k}} \right) \\ $$$${S}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left({e}^{{k}} −\frac{\mathrm{1}}{{e}^{{k}} }\right) \\ $$$${S}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\:\frac{{e}^{\mathrm{2}{k}} −\mathrm{1}}{{e}^{{k}} } \\ $$$$\mathrm{use}\:{L}'{Hopital}'{s}\:{Law} \\ $$$${S}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\:\frac{{e}^{\mathrm{2}{k}} }{\mathrm{2}{e}^{{k}} } \\ $$$${S}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\:\frac{{e}^{{k}} }{\mathrm{2}} \\ $$$$\therefore\:{S}=\infty \\ $$