Question

1/128 times the definite integral from 0 to infinity of y^3 * e^(-y/8)

Answer is 192. I get that we do integration by parts 3 times, but then I do not understand how we end up with 192.

Answer 1

\[\int\limits_{0}^{\inf} = \lim_{t \rightarrow \infty}\int\limits_{0}^{t}\]

lmk if this helps

Answer 2

Thanks, I know how to do that general procedure, but I am having difficulty with this problem specifically

Answer 3

The indefinite integral is:

\[-8e^{-y/8}(y^3 + 24y^2 + 384y + 3072)\]
thanks to wolfram.

to take the limit as y->infinity, are you familiar with L'Hopital's Rule?

Answer 4

Doesn't that just involve taking the derivative until it is simplified?

Answer 5

If you could do it out for me, that would be great

Answer 6

yes. just solved it and doesn't quite work

Answer 7

actually it should. glad i pasted it in a notepad:

pasting:

it is for when the limit of f(x)/g(x) = 0/0 or infinity/infinity:
\[\lim_{x -> \infty}\frac{ f(x) }{ g(x) } = \lim_{x -> \infty}\frac{ d/dx(f(x)) }{ d/dx(g(x)) }\]
not to be confused with quotient rule.
let's put it in the form f(x)/g(x) , limit is clearly infinity/infinity:
\[-8 \lim_{y -> \infty} \frac{ y^3 + 24y^2 + 384y + 3072 }{ e^{y/8} }\]

we derive numerator and denomninator respetively until it is no longer 0/0 or infty/infty

ends up being:\[-8 \lim_{y -> \infty}\frac{ 6 }{ -\frac{ 1 }{ 512 } e^{y/8}} = 0\]


so the entire infinity bound is 0.

plugging in zero leaves you only with -(-8)*(3071)*(1/128) = 192

Answer 8

3072** not 3071 lol

Answer 9

Thank you so much :)

Answer 10

glad i could help ^_^

Answer 11

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