Calculus 2 - Improper integrals determining if they converge.

Question

The integral is from 2 to ∞ ∫ x^2/√(x^7)-1

Answer 1

Notice that ∫x²/√(x⁷-1) dx resembles the form ∫x²/√(x⁷) dx:

∫x²/√(x⁷) dx = ∫x²/x^7/2dx = ∫x^-3/2dx = ∫1/x^3/2dx

Since 3/2>1, we can say the integral ∫1/x^3/2dx converges by the p-test.

Integral Comparison Test

0≤f≤g

If ∫[a,∞] g converges, then ∫[a,∞] f converges

If ∫[a,∞] f diverges, then ∫[a,∞] g diverges

We can write f≤g as x²/√(x⁷-1) ≤ 1/x^3/2

Because we already determined that ∫[2,∞] 1/x^3/2dx converges, then by the integral comparison test, ∫[2,∞] x²/√(x⁷-1) dx also converges!

Hope this helped!

Answer 2

Unfortunately, the previous solution needs some corrections.

Actually, x²/√(x⁷-1) > x²/√x⁷ = 1/x^3/2, so by this way Integral Comparison Test doesn't work.

Small correction.

If x≥2, then x⁷>2,

x⁷/2 > 1,

x⁷-1 > x⁷-x⁷/2 = x⁷/2

x²/√(x⁷-1) <√ x²/√(x⁷/2) = √2 / x^3/2

Now √∫[2,∞] √2/x^3/2dx converges and Integral Comparison Test works.

In addition, we can calculate this integral in the site www.wolframalpha.com

Just enter "integrate x²/√(x⁷-1) from 2 to inf"

The site did not find the antiderivative, but provide the numeric value 1.41458 (very close to √2, but likely it is different from √2)