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

Question

Emma Aldridge · Accepted Answer

Please see below.

#int_0^1 x^(-0.9) dx = lim_(ararr0^+) int_a^1 x^(-0.9) dx# # = lim_(ararr0^+) [10x^0.1]_a^1# # = 10 - lim_(ararr0^+) 10a^0.1# # = 10#

William Abdalla · Answer

undefined 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.