# How do you test the improper integral #int(x-1)^(-2/3)dx# from #[0,1]# and evaluate if possible?

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To test the improper integral ( \int_{0}^{1} (x-1)^{-2/3} dx ), we need to examine whether it converges or diverges.

Firstly, we'll note that the integrand ( (x-1)^{-2/3} ) approaches infinity as ( x ) approaches 1 from the left side. Hence, the integral is improper at the upper limit of integration, ( x = 1 ).

To test for convergence, we'll consider the limit:

[ \lim_{a \to 1^-} \int_{0}^{a} (x-1)^{-2/3} dx ]

We can then evaluate this limit and check if it exists and is finite. If it does, the integral converges; otherwise, it diverges.

To evaluate the integral, we can use the substitution method. Let ( u = x - 1 ), then ( du = dx ).

[ \int (x-1)^{-2/3} dx = \int u^{-2/3} du ]

Now, integrating ( u^{-2/3} ) with respect to ( u ) yields:

[ \frac{u^{1/3}}{1/3} + C = 3u^{1/3} + C ]

Where ( C ) is the constant of integration.

Substituting back ( x - 1 ) for ( u ):

[ = 3(x - 1)^{1/3} + C ]

Now, we can evaluate the improper integral by taking the limit:

[ \lim_{a \to 1^-} 3(x - 1)^{1/3} \Bigg|_{0}^{a} ]

[ = \lim_{a \to 1^-} 3(a - 1)^{1/3} - 3(0 - 1)^{1/3} ]

[ = \lim_{a \to 1^-} 3(a - 1)^{1/3} - 3(-1)^{1/3} ]

[ = 3(0) - 3(-1)^{1/3} ]

[ = -3(-1)^{1/3} ]

Since ( (-1)^{1/3} ) is a complex number, the integral diverges.

Therefore, the improper integral ( \int_{0}^{1} (x-1)^{-2/3} dx ) diverges.

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