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Question Number 157197 by MathSh last updated on 20/Oct/21

Answered by TheSupreme last updated on 20/Oct/21

$$I=\frac{2}{\sqrt{\pi }}{\int }_0^z{e}^{-x^2}dx=0.4$$$$I^2=\frac{4}{\pi }{\int }_0^z{e}^{-x^2-y^2}dxdy=0.16$$$$polar$$$$x=\rho cos\left(\theta \right)$$$$y=\rho sin\left(\theta \right)$$$$det\left(J\right)=\rho$$$$0<\theta <2\pi$$$$0<\rho <z$$$$I^2=\frac{4}{\pi }{\int }_0^z{\int }_0^{2\pi }\rho e^{-{\rho }^2}d\rho d\theta =8{\int }_0^z\rho e^{-{\rho }^2}d\rho =$$$$=8{[-\frac{1}{2}{e}^{-{\rho }^2}]}_0^z=$$$$=4(1-e^{-z^2})=0.16$$$$e^{-z^2}=0.96$$$$-z^2=ln(0.96)$$$$z=\sqrt{-ln(0.96)}$$$$ingen$$$$z=\sqrt{-ln(1-\frac{A^2}{4})}$$

Commented by MathSh last updated on 21/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{dear}\mathrm{Ser}$$

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