Hi, i have a problem which is confusing me(adsbygoogle = window.adsbygoogle || []).push({});

Question:

Given that

[tex]\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]

What is

(i) [tex]\int_{0}^{\infty}e^{-ax^2} x^2 dx[/tex]

(ii) [tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx[/tex]

(iii) [tex]\int_{0}^{\infty}e^{-ax^2} x^4 dx[/tex]

It tells me to use differentiation for (i) and (iii) with respect to the a paramter, and integration by parts for (ii)

I tried (ii):

[tex]u=x^3[/tex]

[tex]\frac{du}{dx} = 3x^2[/tex]

[tex]\frac{dv}{dx} = e^{-ax^2}[/tex]

[tex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]

[tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \int_{0}^{\infty}\frac{3x^2 \sqrt{\pi}}{2\sqrt{a}}[/tex]

[tex]= \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} = 0[/tex]

I know limits should be in that last bit and the bit before but i thought it wouldn't matter as both parts are the same, but then i thought i can't say [itex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/itex] anyway because the given expression wasn't general it was between limits, or does it not matter because i'm using the same limits?

I don't know what the question means when it tells me to "use differentiation for (i) and (iii) with respect to the a paramter" how will that help me get to a solution?

Thanks in advance

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# Homework Help: Integrating e^-ax^2

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