How do you test the improper integral #int x^(-1/3)dx# from #[-1,0]# and evaluate if possible?

Question

Luke Alvarez · Accepted Answer

Use #int x^(-1/3) dx = 3/2x^(2/3)#.

#int_-1^0 x^(-1/3) dx = lim_(brarr0^-)int_-1^b x^(-1/3) dx# # = lim_(brarr0^-) [3/2x^(2/3)]_-1^b# # = lim_(brarr0^-) 3/2 b^(2/3) - 3/2(-1)^(2/3)# # = 0 - 3/2 = -3/2#

Adam Adkins · Answer

undefined To test the improper integral $ \int_{-1}^{0} x^{-1/3} dx $ from $-1$ to $0$, you need to check if the integral converges or diverges.

First, note that the function $x^{-1/3}$ is not defined at $x = 0$. Therefore, you need to evaluate the limit of the integral as it approaches $0$ from the left:

$$ \lim_{a \to 0^-} \int_{-1}^{a} x^{-1/3} dx $$

To evaluate this limit, integrate $x^{-1/3}$ from $-1$ to $a$ and then take the limit as $a$ approaches $0$ from the left.

$$ \lim_{a \to 0^-} \left[ \frac{3x^{2/3}}{2} \right]_{-1}^{a} $$

$$ = \lim_{a \to 0^-} \left( \frac{3a^{2/3}}{2} - \frac{3}{2} \right) $$

$$ = \frac{3}{2} \left( \lim_{a \to 0^-} a^{2/3} - 1 \right) $$

As $a$ approaches $0$ from the left, $a^{2/3}$ approaches $0$ and the limit becomes $-\frac{3}{2}$.

Since the limit exists and is finite, the integral $ \int_{-1}^{0} x^{-1/3} dx $ converges. Therefore, it is possible to evaluate it.