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I need help with the following integral. I used integration by parts to get started but I have no idea where to go from there. If someone could please show me step by step how to solve this, I would be most appreciative!

$$ \int_0^\infty 4y^{2}e^{-2y}dy\,. $$

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Integrate it by parts twice –  M. Strochyk Feb 11 at 16:45
    
Thank you all for the thoughtful answers! –  Jack T Feb 11 at 22:17
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5 Answers

up vote 2 down vote accepted

Hint:

Use integration by parts, twice.

Start with $$u = 4y^2\implies du = 8y\,dy\\dv = e^{-2y} dy\implies v = -\frac 12e^{-2y}$$ Where the integral evaluates to $$uv - \int v\,du$$ That gives us, to start, $$4y^2\cdot -\frac 12e^{-2y} - \int -\frac 12 e^{-2y}(8y\,dy)\\ = -2y^2 e^{-2y} + 4\int ye^{-2y}\,dy$$

Now, one more application of integration by parts, and you're there.

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Integration by parts is the approach you should take in a standard calculus course. We mention an alternative approach. It works, but there is a substantial risk it will be considered unsuitable.

We guess that there is an antiderivative of the shape $(ay^2+by+c)e^{-2y}$. Differentiate. We get $$\left((-2)(ay^2+by+c)+(2ay+b)\right)e^{-2y}.\tag{1}$$ We want (1) to be identically equal to $4y^2e^{-2y}$. So we want $$(-2)(ay^2+by+c) +(2ay+b)=4y^2,\tag{2}$$ identically. There are now various ways to find $a$, $b$, and $c$. For example, the coefficient of $y^2$ on the left is $-2a$, and on the right it is $4$, so $a=-2$. The coefficient of $y$ on the left is $-2b+2a$, on the right it is $0$, so $b=a=-2$. Finally, by comparing constant terms we find that $c=-1$.

Thus $(-2y^2-2y-1)e^{-2y}$ is an antiderivative of $4y^2e^{-2y}$.

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This is often dismissed as an option. Good answer, +1 –  Patrick Da Silva Feb 11 at 17:50
    
Whether dismissed is field dependent. In other contexts, the procedure even has a name. –  André Nicolas Feb 11 at 17:54
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Substitute $2y=x$ to get $\int_0^\infty x^2 e^{-x}dx$ then apply integration by parts

$\int_0^\infty x^2e^{-x}dx=-\int_0^\infty x^2de^{-x}$

$=-x^2e^{-x}|_0^\infty +\int_0^\infty e^{-x}dx^2$

$=0e^0-\lim_{x \to \infty} \frac{x^2}{e^x}+2\int_0 ^\infty xe^{-x}dx$

$=2\int_0^\infty xe^{-x}dx=-2\int_0^\infty xde^{-x}=-2xe^{-x}|_0^\infty+2\int_0^\infty e^{-x}dx$

$=-e^{-x}|_0^\infty=0+1=1.$

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If you know some probability, you may recognize that an Exponential$(\lambda)$ random variable has the density $f(x) = \lambda e^{-\lambda x}$ for $x\geq 0$ and $0$ otherwise. The integral $\int_0^\infty x^2 f(x) dx$ is the second moment of an exponential random variable, which is the $variance + (mean)^2$. The mean of an exponential is $\lambda^{-1}$ and variance is $\lambda^{-2}$, so the second moment of an exponential is $\frac{2}{\lambda^2}$.

Taking $\lambda = 2$, we see that the second moment of an Exponential$(2)$ random variable is half of the desired integral.

Of course, you need to have shown a related integral in some other way at some point (e.g. integration by parts twice, or characteristic functions) , probably, but it is a handy way of doing this sort of computation quickly.

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If you want a shorter answer to your question, apart from the other good ones, you may consider this approach. Notice that the integrand and the limits of integration resembles a lot to the definition of the Laplace transform for a function defined in $y>0^+$ (depending on the author). So, indeed:

$$I = \int^{\infty}_0 4y^2 \, e^{-2y} \, dy = \int^{\infty}_0 4y^2 \, e^{-sy}|_{s=2} \, dy = \mathcal{L}_s[\, 4y^2\,]|_{s=2} = \left.\frac{8}{s^3}\right|_{s=2} = 1.$$

I hope this may be useful to you or interesting for everyone else.

Cheers!

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