33, 34, and 35 evaluate the integral, given that\n int _ { 0 } ^ { infty } e ^ { - x ^ { 2 } } d x = \frac {…

Explanation:

Step1: Use substitution for problem 33

Let ( t=\frac{x}{\sqrt{2}} ), then ( x = \sqrt{2}t ) and ( dx=\sqrt{2}dt ). When ( x = 0,t = 0 ); when ( x\rightarrow\infty,t\rightarrow\infty ). The integral ( \int_{0}^{\infty}e^{-\frac{x^{2}}{2}}dx=\int_{0}^{\infty}e^{-t^{2}}\sqrt{2}dt ) Since ( \int_{0}^{\infty}e^{-x^{2}}dx=\frac{1}{2}\sqrt{\pi} ), then ( \int_{0}^{\infty}e^{-t^{2}}\sqrt{2}dt=\sqrt{2}\times\frac{1}{2}\sqrt{\pi}=\frac{\sqrt{2\pi}}{2} )

Step2: Use integration - by - parts for problem 34

Let ( u = x ), ( dv=x e^{-x^{2}}dx ). Then ( du = dx ), ( v=-\frac{1}{2}e^{-x^{2}} ) By integration - by - parts formula ( \int_{0}^{\infty}x^{2}e^{-x^{2}}dx=\left[- \frac{1}{2}xe^{-x^{2}}\right]{0}^{\infty}+\frac{1}{2}\int{0}^{\infty}e^{-x^{2}}dx ) (\lim_{x\rightarrow\infty}-\frac{1}{2}xe^{-x^{2}}=\lim_{x\rightarrow\infty}-\frac{x}{2e^{x^{2}}}), using L'Hopital's rule (differentiate numerator and denominator), (\lim_{x\rightarrow\infty}-\frac{1}{4xe^{x^{2}}}=0), and when ( x = 0,-\frac{1}{2}xe^{-x^{2}} = 0 ) Since ( \int_{0}^{\infty}e^{-x^{2}}dx=\frac{1}{2}\sqrt{\pi} ), then ( \int_{0}^{\infty}x^{2}e^{-x^{2}}dx=\frac{1}{4}\sqrt{\pi} )

Step3: Use substitution for problem 35

Let ( u=\sqrt{x} ), then ( x = u^{2} ), ( dx = 2udu ). When ( x = 0,u = 0 ); when ( x\rightarrow\infty,u\rightarrow\infty ) The integral ( \int_{0}^{\infty}\sqrt{x}e^{-x}dx=\int_{0}^{\infty}u e^{-u^{2}}\times2udu=2\int_{0}^{\infty}u^{2}e^{-u^{2}}du ) From problem 34, ( \int_{0}^{\infty}u^{2}e^{-u^{2}}du=\frac{1}{4}\sqrt{\pi} ), so ( 2\int_{0}^{\infty}u^{2}e^{-u^{2}}du=\frac{1}{2}\sqrt{\pi} )

Answer:

33. (\frac{\sqrt{2\pi}}{2})
34. (\frac{1}{4}\sqrt{\pi})
35. (\frac{1}{2}\sqrt{\pi})