Question

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

Find the value of integral from 0 to infinity of [x*f(x)]dx given the fact that the integral from 0 to infinity of (e^(-x^2))dx = root pi / 2

Answer 1

`\int^(\infty)_(0)xf(x)dx=(\pi)/(4)`

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=-(1)/(2)\int e^(a)da`

`v=-(1)/(2)e^{-x^(2)}`

Using this in (1) we have:

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

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

`\int^(\infty)_(0)xf(x)dx=(\pi)/(4)`

I hope it helps you!