How do I integrate 4x^2/(x^3-8) from 0 to 8
It is an improper integral and I know it diverges but I don't know why
i found u=x^3-8
but am getting stuck after that
How do I integrate 4x^2/(x^3-8) from 0 to 8
It is an improper integral and I know it diverges but I don't know why
i found u=x^3-8
but am getting stuck after that
Or you can use a u substitution that you tried.
We have u = x^{3} - 8, and du = 3x^{2}dx so
∫ [4x^{2}/(x^{3 }- 8)] dx = (4/3) ∫ du/u = (4/3) ln|u| + C = (4/3) ln|x^{3 }- 8| + C
from 0 to 8 it diverges since at x = 2 you get (4/3) ln 0 → ∞ as said by Robert.
You can tell it diverges even before doing the integral, because at x = 2, the denominator of 4x^{2}/(x^{3} - 8) is zero. There's a vertical asymptote at x = 2 (and that makes it difficult to find the area under the curve!).
Remember that that's the point of integration - to find the area under a curve, and you can't find the area under a vertical asymptote. Looking at a graph of the function may help to see this:
∫4x^2/(x^3-8) dx
= ∫(4/3)/(x^3-8) d(x^3-8)
= (4/3)ln|x^3-8|, which approaches infinity as x --> 2.
So, the integral diverges.
Comments
Wow thanks for your help
That actually makes sense