Question Number 157197 by MathSh last updated on 20/Oct/21
Answered by TheSupreme last updated on 20/Oct/21
![I=(2/( (√π)))∫_0 ^z e^(−x^2 ) dx=0.4 I^2 =(4/π)∫_0 ^z e^(−x^2 −y^2 ) dxdy=0.16 polar x=ρ cos(θ) y= ρ sin(θ) det(J)=ρ 0<θ<2π 0<ρ<z I^2 =(4/π)∫_0 ^z ∫_0 ^(2π) ρe^(−ρ^2 ) dρdθ=8 ∫_0 ^z ρe^(−ρ^2 ) dρ= =8[−(1/2)e^(−ρ^2 ) ]_0 ^z = =4(1−e^(−z^2 ) )=0.16 e^(−z^2 ) =0.96 −z^2 =ln(0.96) z=(√(−ln(0.96))) in gen z=(√(−ln(1−(A^2 /4))))](https://www.tinkutara.com/question/Q157201.png)
$${I}=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{z}} {e}^{−{x}^{\mathrm{2}} } {dx}=\mathrm{0}.\mathrm{4} \\ $$$${I}^{\mathrm{2}} =\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{{z}} {e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}=\mathrm{0}.\mathrm{16} \\ $$$${polar} \\ $$$${x}=\rho\:{cos}\left(\theta\right) \\ $$$${y}=\:\rho\:{sin}\left(\theta\right) \\ $$$${det}\left({J}\right)=\rho \\ $$$$\mathrm{0}<\theta<\mathrm{2}\pi \\ $$$$\mathrm{0}<\rho<{z} \\ $$$${I}^{\mathrm{2}} =\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{{z}} \int_{\mathrm{0}} ^{\mathrm{2}\pi} \rho{e}^{−\rho^{\mathrm{2}} } {d}\rho{d}\theta=\mathrm{8}\:\int_{\mathrm{0}} ^{{z}} \rho{e}^{−\rho^{\mathrm{2}} } {d}\rho= \\ $$$$=\mathrm{8}\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\rho^{\mathrm{2}} } \right]_{\mathrm{0}} ^{{z}} = \\ $$$$=\mathrm{4}\left(\mathrm{1}−{e}^{−{z}^{\mathrm{2}} } \right)=\mathrm{0}.\mathrm{16} \\ $$$${e}^{−{z}^{\mathrm{2}} } =\mathrm{0}.\mathrm{96} \\ $$$$−{z}^{\mathrm{2}} ={ln}\left(\mathrm{0}.\mathrm{96}\right) \\ $$$${z}=\sqrt{−{ln}\left(\mathrm{0}.\mathrm{96}\right)} \\ $$$${in}\:{gen} \\ $$$${z}=\sqrt{−{ln}\left(\mathrm{1}−\frac{{A}^{\mathrm{2}} }{\mathrm{4}}\right)} \\ $$
Commented by MathSh last updated on 21/Oct/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Ser} \\ $$