# How do you test the improper integral #int x^-0.9 dx# from #[0,1]# and evaluate if possible?

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To test the improper integral (\int_{0}^{1} x^{-0.9} dx) from 0 to 1, you need to check if the integral converges or diverges.

The integral (\int_{0}^{1} x^{-0.9} dx) is improper because it involves an infinite discontinuity at (x = 0).

To evaluate the improper integral, integrate the function (x^{-0.9}) from (x = \epsilon) to (x = 1) where (\epsilon) is a small positive number approaching 0, then take the limit as (\epsilon) approaches 0.

[ \lim_{{\epsilon \to 0^+}} \int_{\epsilon}^{1} x^{-0.9} dx ]

If this limit exists and is finite, then the improper integral converges. If the limit does not exist or is infinite, then the improper integral diverges.

Evaluate the integral:

[ \int_{\epsilon}^{1} x^{-0.9} dx = \frac{x^{0.1}}{0.1} \bigg|_{\epsilon}^{1} ]

[ = \frac{1}{0.1} - \frac{\epsilon^{0.1}}{0.1} ]

[ = 10 - \frac{1}{0.1} \epsilon^{0.1} ]

Now take the limit as (\epsilon) approaches 0:

[ \lim_{{\epsilon \to 0^+}} \left(10 - \frac{1}{0.1} \epsilon^{0.1}\right) ]

[ = 10 - \frac{1}{0.1} \cdot 0 ]

[ = 10 ]

Since the limit is finite, the improper integral (\int_{0}^{1} x^{-0.9} dx) converges, and its value is 10.

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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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