How do you evaluate the integral #int 1/(x-1)^(2/3)dx# from 0 to 2?

Answer 1

To evaluate #int_0^2 1/(x-1)^(2/3)dx#, the integrand, i.e.,

#1/(x-1)^(2/3)# has to be continuous over the interval #[0,2]#.

Here, it is not so at #x=1 in [0,2]#.

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Answer 2

The integrand is not defined at #x = 1#, so this is an improper integral.

We need to break this into two improper integrals and try to evaluate each of them. If both integrals converge, then we can add the values to get the integral on #[0,2]#
#int_0^2 1/(x-1)^(2/3) dx = int_0^1 1/(x-1)^(2/3) dx+int_1^2 1/(x-1)^(2/3) dx#

provided that both integrals on the right converge,

# = lim_(brarr1^-)int_0^b 1/(x-1)^(2/3) dx + lim_(ararr1^+)int_a^2 1/(x-1)^(2/3) dx#

provided that both integrals on the right converge.

#lim_(brarr1^-)int_0^b 1/(x-1)^(2/3) dx ={: lim_(brarr1^-) 3(x-1)^(1/3) ]_0^b = 3#

and

#lim_(ararr1^+)int_a^1 1/(x-1)^(2/3) dx ={: lim_(ararr1^+) 3(x-1)^(1/3) ]_a^2 = 3#

We conclude:

#int_0^2 1/(x-1)^(2/3) dx = int_0^1 1/(x-1)^(2/3) dx+int_1^2 1/(x-1)^(2/3) dx#
# = 3+3=6#
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Answer 3

To evaluate the integral (\int_{0}^{2} \frac{1}{(x-1)^{\frac{2}{3}}} dx), we first need to recognize that this integral involves an improper integral due to the singularity at (x = 1). To handle this, we can make a substitution to transform the integral into a standard form.

Let (u = x - 1), then (du = dx). Also, when (x = 0), (u = -1), and when (x = 2), (u = 1).

Substituting these into the integral, we get:

[ \int_{-1}^{1} \frac{1}{u^{\frac{2}{3}}} du ]

This integral can be evaluated using the power rule for integration.

[ = \left[\frac{3}{1-\frac{2}{3}} u^{1-\frac{2}{3}}\right]_{-1}^{1} ]

[ = \left[ \frac{3}{\frac{1}{3}}u^{\frac{1}{3}} \right]_{-1}^{1} ]

[ = 9\left(\left(1\right)^{\frac{1}{3}} - \left(-1\right)^{\frac{1}{3}}\right) ]

[ = 9\left(1 - (-1)\right) ]

[ = 9(2) ]

[ = 18 ]

Therefore, the value of the integral (\int_{0}^{2} \frac{1}{(x-1)^{\frac{2}{3}}} dx) is (18).

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Answer from HIX Tutor

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.

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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