Answer to Question #139465 in Calculus for Kembot Lav

Question

integral of e^(-x^2)

Answer 1

$Integral \ of \ {e^{-x^2}}$

Firstly, set the integral to be equal to I

$I = \int{e^{-x^2}dx}$

Squaring both sides,

$I^2 = {(\int{e^{-x^2}dx})}^2 =(\int{e^{-x^2}dx})(\int{e^{-x^2}dx})$

Since x is just a dummy variable, we can write,

$I^2=(\int{e^{-x^2}dx})(\int{e^{-y^2}dy})$

Combining, we have,

$I^2=\int^{\infin}_{-\infin}(\int^{\infin}_{-\infin}{e^{-(x^2+y^2)} \ dx)dy}$

$I^2=\int^{\infin}_{-\infin}\int^{\infin}_{-\infin}{e^{-(x^2+y^2)} \ dxdy}$

Converting from Cartesian Coordinates (x,y) to Polar Coordinates (r,θ)

$(r^2 =x^2 + y^2)$

$($$dxdy =dA = rdrd\,\theta)$

$\therefore I^2 =\iint{e^{-r^2} rdrd\,\theta}$

$I^2 =\int^{2\pi}_{0}(\int^{\infin}_{0}{e^{-r^2} r\,dr)d\,\theta}$

$I^2 =\int^{2\pi}_{0}(\int^{\infin}_{0}{re^{-r^2} dr)d\,\theta}$

$I^2 =2\pi(\int^{\infin}_{0}{re^{-r^2}dr})$

$I^2 =2\pi\int^{\infin}_{0}{re^{-r^2} dr}$

Using u substitution,

$u = r^2$

$\dfrac{du}{dr} = 2r$

$du = 2r\,dr$

$\therefore I^2 =2\pi\int^{\infin}_{u =0}{re^{-u}dr}$

Multiplying both sides of the equation by 2,

$2I^2 =2\pi\int^{\infin}_{u =0}{2re^{-u}dr}$

But du = 2rdr

$\therefore 2I^2 =2\pi\int^{\infin}_{u =0}{e^{-u}du}$

$2I^2 =2\pi[e^{0}-e^{-\infin}]$

$2I^2 =2\pi(1-0)$

$2I^2 =2\pi$

$I^2 =\pi$

$I =\sqrt\pi$

$\therefore \ The \ integral \ of \ e^{−x^2} \ is \ \sqrt{\pi}$