How do you test the improper integral $\int (x^{- 2} - x^{- 3}) d x$ $int (x^-2-x^-3)dx$ from $[3, \infty)$ $[3,oo)$ and evaluate if possible?

Question

1503 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-22T22:46:08

$\int_{3}^{\infty} (x^{- 2} - x^{- 3}) d x = \frac{5}{18}$ $int_3^oo (x^-2-x^-3)dx = 5/18$

We have:

$int_3^oo (x^-2-x^-3)dx = lim_(u->oo) int_3^u (x^-2-x^-3)dx$

Using the power rule:

$int_3^u (x^-2-x^-3)dx = [-x^-1 +x^-2/2]_3^u = 1/(2u^2)-1/u +1/3-1/18 = 1/u(1/(2u)-1) +5/18$

So:

$int_3^oo (x^-2-x^-3)dx = lim_(u->oo) 1/u(1/(2u)-1) +5/18 = 0*(-1)+5/18 = 5/18$

How do you test the improper integral int (x^-2-x^-3)dx∫(x−2−x−3)dxint (x^-2-x^-3)dx from [3,oo)[3,∞)[3,oo) and evaluate if possible?