Question

how do I solve this improper integral question? question in the comments

Answer 1

I know the answer but I don't know how I can get the answer that is not 0 or infinity with this integral

Answer 2

you have to use this:
\[\int\limits_{0}^{\infty} = \lim_{t ->\infty}\int\limits_{0}^{t}\]

let me know if you get the answer. otherwise i'll solve it :)

Answer 3

hi. I got stuck at limit t -> inifinity pi integral 0 to t 1/(0.64x^2+2.88x+3.24)dx
I don't know how to solve that integral without using complete squares

Answer 4

you shouldn't have expanded the bracket. let u = 0.8x + 1.8; du = 0.8 dx
so it's the integral of u^(-2) du

\[\pi \lim_{t -> \infty}\int\limits_{0}^{t}\frac{ dx }{ (0.8x + 1.8)^2 } = \frac{ -\pi }{ 0.8 }\frac{ 1 }{ (0.8x + 1.8) } = \frac{ -\pi }{ 0.8 }\left[ \left( \lim_{t - > \infty}\frac{ 1 }{ 0.8t } \right) - \frac{ 1 }{ 0.8(0) - 1.8 } \right]\]

Answer 5

note: it should be [0.8(0) + 1.8] in denominator of last term obviously. i put minus by accident. and 1/(lim->infinity) = 0

also, there is nothing after the cutoff. just simplify and that is all.