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

Note that:

So:

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To test the improper integral ( \int_{2}^{\infty} x^{-2} dx ), we need to find out whether it converges or diverges.

To evaluate the integral, we integrate ( x^{-2} ) with respect to ( x ) from ( x = 2 ) to ( x = \infty ).

[ \int_{2}^{\infty} x^{-2} dx = \lim_{a \to \infty} \int_{2}^{a} x^{-2} dx ]

[ = \lim_{a \to \infty} \left[-\frac{1}{x}\right]_{2}^{a} ]

[ = \lim_{a \to \infty} \left(-\frac{1}{a} + \frac{1}{2}\right) ]

[ = \frac{1}{2} ]

Therefore, the integral converges and its value is ( \frac{1}{2} ).

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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.

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.

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.

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