## Thomas' Calculus 13th Edition

A) $\frac{\sqrt{\pi}}{2}$ B) 1
A) $I^2 =\int^{\infty}_0\int^{\infty}_0 e^{-(x^2+y^2)}dxdy$ =$\int^{\pi/2}_ 0 \int^{\infty}_0(e^{-r^2})rdrd\theta$ =$\int^{\pi/2}_0 [\lim\limits_{b \to \infty}\int^b_ re^{-r^2}]d\theta$ =$-\frac{1}{2} \int^{\pi/2}_0 \lim\limits_{b \to \infty}(e^{-b^2}-1)d\theta$ =$\frac{1}{2} \int^{\pi/2}_0 d\theta =\frac{\pi}{4}$ Thus: $I=\frac{\sqrt{\pi}}{2}$ --- B) $\lim\limits_{x \to \infty} \int^x_0 \frac{2e^{-t^2}}{\sqrt{\pi}}dt$ =$\frac{2}{\sqrt{\pi}}\int^\infty_0 e^{\it^2} dt$ =$(\frac{2}{\sqrt{\pi}}(\frac{\sqrt{\pi}}{2}))=1$