Question

Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.

Answer 1

Answer:

$\frac{\sqrt{\pi}}{4}$

Step-by-step explanation:

You are going to integrate the following function:

$g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}$  (1)

furthermore, you know that:

$\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}$

lets call to this integral, the integral Io.

for a general form of I you have In:

$I_n=\int_0^{\infty}x^ne^{-ax^2}dx$

furthermore you use the fact that:

$I_n=-\frac{\partial I_{n-2}}{\partial a}$

by using this last expression in an iterative way you obtain the following:

$\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}}$ (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

$\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}$

Answer 2

Answer:

$\int^{\infty}_{0}xf(x)dx=\frac{\pi}{4}$  

Step-by-step explanation:

We need to find the integrate of:

$\int^{\infty}_{0}xf(x)dx$

Let's use the integration by parts rule.

$\int^{\infty}_{0}xf(x)dx=u*v-\int vdu$ (1)

$u=x$ and $du=dx$

$dv=f(x)dx$ and $v=\int f(x)dx$

$f(x)=xe^{-x^{2}}$

$v=\int xe^{-x^{2}}dx$ if we use change variable we can solve it, we can do $a=-x^{2}$ then $da=-2xdx$

So we have:

$v=-\frac{1}{2}\int e^{a}da$

$v=-\frac{1}{2}e^{-x^{2}}$

Using this in (1) we have:

$\int^{\infty}_{0}xf(x)dx=(-\frac{1}{2}x*e^{-x^{2}})|^{\infty}_{0}-\int^{\infty}_{0} (-\frac{1}{2}e^{-x^{2}})dx$

The used criteria to make the first term zero is because the exponential tends to zero faster than the x tends to infinity.    

$\int^{\infty}_{0}xf(x)dx=0+\frac{1}{2}\int^{\infty}_{0} e^{-x^{2}}dx$  

We know that the integral from 0 to infinity of $e^{-x^{2}}=\frac{\sqrt{\pi}}{2}$, hence:  $\int^{\infty}_{0}xf(x)dx=0+\frac{1}{2}\frac{\sqrt{\pi}}{2}$

$\int^{\infty}_{0}xf(x)dx=\frac{\pi}{4}$  

I hope it helps you!