How do you test the improper integral #int x^(-2/3)dx# from #[-1,1]# and evaluate if possible?

Question

Elijah Abram · Accepted Answer

The answer is #=6#

The improper integral is

#int_-1^1x^(-2/3dx)#

There is an undefined point when #x=0#

Therefore,

#int_-1^1x^(-2/3dx)=lim_(p->0)int_-1^px^(-2/3)dx+lim_(p->0)int_p^1x^(-2/3)dx#

#lim_(p->0)int_-1^px^(-2/3)dx=lim_(p->0)[3x^(1/3)]_-1^p#

#=lim_(p->0)(3p^(1/3)+3)#

#=3#

#lim_(p->0)int_p^1x^(-2/3)dx=lim_(p->0)[3x^(1/3)]_p^1#

#=lim_(p->0)(3-3p^(1/3))#

#=3#

Finally,

#int_-1^1x^(-2/3dx)=3+3=6#

William Abdalla · Answer

undefined To test the improper integral $ \int_{-1}^{1} x^{-2/3} \, dx $ and evaluate it, follow these steps:

1. Identify if there are any discontinuities or singularities within the interval of integration.
2. Determine if the function $ x^{-2/3} $ is integrable over the interval [-1, 1].
3. If there are discontinuities or singularities, split the integral into multiple parts around those points.
4. Integrate each part separately, and then evaluate the limit as the discontinuity approaches the endpoint of the interval.

Given the function $ f(x) = x^{-2/3} $, it has a singularity at $ x = 0 $. Therefore, the interval [-1, 1] includes this singularity.

To evaluate the integral, split it into two parts:

1. $ \int_{-1}^{0} x^{-2/3} \, dx $
2. $ \int_{0}^{1} x^{-2/3} \, dx $

For each part:

$$ \int x^{-2/3} \, dx = \frac{3x^{1/3}}{1/3} + C = 3x^{1/3} + C $$

Now, evaluate each part separately:

1. For $ \int_{-1}^{0} x^{-2/3} \, dx $:
$$ = [3x^{1/3}]_{-1}^{0} $$
$$ = 3(0)^{1/3} - 3(-1)^{1/3} $$
$$ = -3 - 0 $$
$$ = -3 $$

2. For $ \int_{0}^{1} x^{-2/3} \, dx $:
$$ = [3x^{1/3}]_{0}^{1} $$
$$ = 3(1)^{1/3} - 3(0)^{1/3} $$
$$ = 3 - 0 $$
$$ = 3 $$

Combine the results:

$$ \int_{-1}^{1} x^{-2/3} \, dx = \int_{-1}^{0} x^{-2/3} \, dx + \int_{0}^{1} x^{-2/3} \, dx $$
$$ = -3 + 3 $$
$$ = 0 $$

Therefore, the value of the improper integral $ \int_{-1}^{1} x^{-2/3} \, dx $ is 0.

Cameron Alexander · Answer

undefined To test the improper integral $ \int_{-1}^{1} x^{-2/3} \, dx $, we first need to check if it converges or diverges. Since the integrand $ x^{-2/3} $ has a singularity at $ x = 0 $ within the interval of integration, the integral is improper.

To evaluate this improper integral, we split it into two separate integrals:

$$ \int_{-1}^{1} x^{-2/3} \, dx = \lim_{a \to 0^-} \int_{-1}^{a} x^{-2/3} \, dx + \lim_{b \to 0^+} \int_{b}^{1} x^{-2/3} \, dx $$

Then, we integrate each part individually:

$$ \lim_{a \to 0^-} \int_{-1}^{a} x^{-2/3} \, dx = \lim_{a \to 0^-} \left[ \frac{3x^{1/3}}{1/3} \right]_{-1}^{a} $$
$$ = \lim_{a \to 0^-} \left( 3a^{1/3} - 3(-1)^{1/3} \right) $$
$$ = 3 \lim_{a \to 0^-} a^{1/3} + 3 $$

$$ \lim_{b \to 0^+} \int_{b}^{1} x^{-2/3} \, dx = \lim_{b \to 0^+} \left[ \frac{3x^{1/3}}{1/3} \right]_{b}^{1} $$
$$ = \lim_{b \to 0^+} \left( 3 - 3b^{1/3} \right) $$
$$ = 3 - 3 \lim_{b \to 0^+} b^{1/3} $$

Now, let's evaluate the limits:

$$ \lim_{a \to 0^-} a^{1/3} = 0 $$
$$ \lim_{b \to 0^+} b^{1/3} = 0 $$

Thus, both limits are finite, and the improper integral converges. We can then sum the two parts:

$$ \int_{-1}^{1} x^{-2/3} \, dx = 3 + 3 = 6 $$

So, the value of the improper integral $ \int_{-1}^{1} x^{-2/3} \, dx $ is 6.