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

Answer 1

Christopher Allers

#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#

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Answer 2

Axel Alvarez

To test the improper integral (\int_{3}^{\infty} (x^{-2} - x^{-3}) , dx), first integrate each term separately:

(\int_{3}^{\infty} x^{-2} , dx) and (\int_{3}^{\infty} x^{-3} , dx)

The integral of (x^{-2}) is (\lim_{a \to \infty} [-\frac{1}{x}]{3}^{a} = \lim{a \to \infty} (-\frac{1}{a} + \frac{1}{3})).

Similarly, the integral of (x^{-3}) is (\lim_{a \to \infty} [-\frac{1}{2x^2}]{3}^{a} = \lim{a \to \infty} (-\frac{1}{2a^2} + \frac{1}{18})).

Evaluate these limits to determine if they converge or diverge. If both limits converge, then the integral converges. If one or both limits diverge, the integral diverges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.