Answer:
[tex]\int^{\infty}_{0}xf(x)dx=\frac{\pi}{4}[/tex]
Step-by-step explanation:
We need to find the integrate of:
[tex]\int^{\infty}_{0}xf(x)dx[/tex]
Let's use the integration by parts rule.
[tex]\int^{\infty}_{0}xf(x)dx=u*v-\int vdu[/tex] (1)
[tex]u=x[/tex] and [tex]du=dx[/tex]
[tex]dv=f(x)dx[/tex] and [tex]v=\int f(x)dx[/tex]
[tex]f(x)=xe^{-x^{2}}[/tex]
[tex]v=\int xe^{-x^{2}}dx[/tex] if we use change variable we can solve it, we can do [tex]a=-x^{2}[/tex] then [tex]da=-2xdx[/tex]
So we have:
[tex]v=-\frac{1}{2}\int e^{a}da[/tex]
[tex]v=-\frac{1}{2}e^{-x^{2}}[/tex]
Using this in (1) we have:
[tex]\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[/tex]
The used criteria to make the first term zero is because the exponential tends to zero faster than the x tends to infinity.
[tex]\int^{\infty}_{0}xf(x)dx=0+\frac{1}{2}\int^{\infty}_{0} e^{-x^{2}}dx[/tex]
We know that the integral from 0 to infinity of [tex]e^{-x^{2}}=\frac{\sqrt{\pi}}{2}[/tex], hence: [tex]\int^{\infty}_{0}xf(x)dx=0+\frac{1}{2}\frac{\sqrt{\pi}}{2}[/tex]
[tex]\int^{\infty}_{0}xf(x)dx=\frac{\pi}{4}[/tex]
I hope it helps you!
The definite integral of [tex]f(x) = x\cdot e^{-x^{2}}[/tex] from 0 to [tex]+\infty[/tex] is [tex]\frac{1}{2}[/tex].
The integral given in statement can be solved by integration by substitution:
[tex]\int\limits^{+\infty}_{0} {x\cdot e^{-x^{2}}} \, dx = -\frac{1}{2} \int\limits^{+\infty}_{0} {e^{u}} \, du[/tex], where [tex]u = -x^{2}[/tex] (1)
[tex]\int\limits^{+\infty}_{0} {x\cdot e^{-x^{2}}} \, dx = \frac{1}{2}\cdot e^{-x^{2}}|_{0}^{+\infty}[/tex]
We could notice that [tex]e^{-x^{2} } \to 0[/tex] when [tex]x \to + \infty[/tex]. Then, we have the following result:
[tex]\int\limits^{+\infty}_{0} {x\cdot e^{-x^{2}}} \, dx = \frac{1}{2}\cdot (1-0)[/tex]
[tex]\int\limits^{+\infty}_{0} {x\cdot e^{-x^{2}}} \, dx = \frac{1}{2}[/tex]
The definite integral of [tex]f(x) = x\cdot e^{-x^{2}}[/tex] from 0 to [tex]+\infty[/tex] is [tex]\frac{1}{2}[/tex]. [tex]\blacksquare[/tex]
To learn more on definite integrals, we kindly invite to check this verified question: https://brainly.com/question/22655212
