Question

can anyone help me with Integration of (x^2)*(exp(-a*x)) from 0 to infinity

Answer 1

\[\int\limits_0^\infty x^2e^{-ax}\text dx\]
you will have to use integration by parts twice

\(\int u\text dv=\left. uv\right|-\int v\text du\)

for the first time ,\[u=x^2\qquad\qquad \text dv=e^{-ax}\]\[\text du=2x\text dx\qquad\qquad v=\frac{e^{-ax}}{-a}\],

dont forget the limits

Answer 2

* \(\text dv=e^{-ax}\text dx\)

Answer 3

that i know,but is the final ans infinity or some function in only a....if latter,what is that function?

Answer 4

the final result is a function of a

Answer 5

i m asking that function of a...i m getting infinity,which is clearly wrong

Answer 6

is it just me or does this thing look a lot like laplace formulas :/

Answer 7

\[\int\limits_0^\infty x^2e^{-ax}\text dx=\left.x^2\frac{e^{-ax}}{-a}\right|_0^∞-\int\limits_0^∞2x\frac{e^{-ax}}{-a}\text dx\]
\[=2\int\limits_0^∞x\frac{e^{-ax}}{a}\text dx\]

now use integration by parts again
\(\int p\text dq=\left. pq\right|-\int q\text dp\)

\[p=x\qquad\qquad\text dq=\frac{e^{-ax}}{a}\text dx\]\[\text dp=\text dx\qquad\qquad q= \frac{e^{-ax}}{-a}\]

Answer 8

\[\qquad\qquad\qquad * q=\frac{e^{-ax}}{-a^2}\]

\[2\int\limits_0^∞x\frac{e^{-ax}}{a}\text dx=2\left[\left.x\frac{e^{-ax}}{-a^2}\right|_0^∞-\int\frac{e^{-ax}}{-a^2}\text dx\right]\]
\[\qquad=2\left[\int\frac{e^{-ax}}{a^2}\text dx\right]\]\[=2\left.\frac{e^{-ax}}{a^3}\right|_0^∞\]\[=\frac{2}{a^3}\]

Answer 9

BLAM!

Answer 10

any steps you not sure on, @hartnn?

Answer 11

note \[e^{-∞} =0\]

Answer 12

actually i m not able to read/understand your response....the two responses before blam...

Answer 13

actually i m not able to read/understand your response....the two responses before blam...

Answer 14

all i have done is to integrate by parts twice

Answer 15

can you be more specific ?

Answer 16

got it,thanks so much :)

Answer 17

it all makes sense now/?

Answer 18

but x^2*e^-ax when we put infinity,isn't it infinity*0???

Answer 19

and not 0

Answer 20

\[\left.x^2\frac{e^{-ax}}{-a}\right|_0^∞=\left(\infty^2\frac{e^{-a\times\infty}}{-a}\right)-\left(0^2\frac{e^{-a\times0}}{-a}\right)\]\[\qquad\qquad=\left(\infty\times0\right)-\left(0\times0\right)\]\[\qquad=0-0\]\[=0\]

Answer 21

@lgbasallote is right, the easy way {if you know how}, is to recognise the laplace form

\[\int\limits_0^\infty x^2e^{-ax}\text dx=\mathcal L\{x^2\}(a)=\frac{\Gamma(2+ 1)}{a^{2+1}}=\frac{2!}{a^3}=\frac2{a^3}\]