Question Number 220563 by mathocean1 last updated on 15/May/25
$${Calculate}\:{the}\:{exact}\:{value}\:{of}\:: \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{4}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$
Answered by SdC355 last updated on 15/May/25
$$\mathrm{easy}. \\ $$$$\int\:\:{e}^{−{z}^{\mathrm{2}} } \:\mathrm{d}{z}=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\left({z}\right)+\mathrm{const}. \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{4}} \:{e}^{−{z}^{\mathrm{2}} } \mathrm{d}{z}=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\left(\mathrm{4}\right)−\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\left(\mathrm{0}\right) \\ $$$$\mathrm{erf}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{4}} \:{e}^{−{z}^{\mathrm{2}} } \mathrm{d}{z}=\frac{\sqrt{\pi}}{\mathrm{2}}\:\mathrm{erf}\left(\mathrm{4}\right) \\ $$$$\mathrm{or}\:{I}=\int_{\mathrm{0}} ^{\:\mathrm{4}} {e}^{−{z}^{\mathrm{2}} } \mathrm{d}{z} \\ $$$${I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\mathrm{4}} \int_{\mathrm{0}} ^{\:\mathrm{4}} \:{e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \mathrm{d}{x}\mathrm{d}{y} \\ $$$${x}={r}\mathrm{cos}\left(\theta\right) \\ $$$${y}={r}\mathrm{sin}\left(\theta\right) \\ $$$${J}=\mid\mid\frac{\partial\left({x},{y}\right)}{\partial\left({r},\theta\right)}\mid\mid\:\mathrm{d}{r}\mathrm{d}\theta=\begin{vmatrix}{\mathrm{cos}\left(\theta\right)}&{−{r}\mathrm{sin}\left(\theta\right)}\\{\mathrm{sin}\left(\theta\right)}&{\:\:\:\:{r}\mathrm{cos}\left(\theta\right)}\end{vmatrix}= \\ $$$$\left({r}\mathrm{cos}^{\mathrm{2}} \left(\theta\right)+{r}\mathrm{sin}^{\mathrm{2}} \left(\theta\right)\right)\mathrm{d}{r}\mathrm{d}\theta={r}\mathrm{d}{r}\mathrm{d}\theta \\ $$$$\int\int_{\:\mathcal{D}} \:{re}^{−{r}^{\mathrm{2}} } \mathrm{d}{r}\mathrm{d}\theta=\int_{\mathrm{0}} ^{\:\pi} \int_{\mathrm{0}} ^{\:\mathrm{4}} \:{re}^{−{r}^{\mathrm{2}} } \mathrm{d}{r}\mathrm{d}\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\sqrt{\int\int_{\:\mathcal{D}} {re}^{−{r}^{\mathrm{2}} } \mathrm{d}{r}\mathrm{d}\theta}\overset{\mathrm{app}} {\approx}\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\mathrm{erf}\left(\mathrm{4}\right)\overset{\mathrm{approximatly}} {\approx}\mathrm{0}.\mathrm{99999998458274}…….\: \\ $$$$\mathrm{So},\:\mathrm{we}\:\mathrm{can}\:\mathrm{consider}\:\mathrm{erf}\left(\mathrm{4}\right)\:\mathrm{as}\:\mathrm{almost}\:\mathrm{1}. \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}erf}\left({x}\right)=\mathrm{1}. \\ $$
Commented by mathocean1 last updated on 15/May/25
$${thank}\:{you}\:{sir} \\ $$