# How do you evaluate the integral #int 1/(x-2)^(2/3)dx# from 1 to 4?

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To evaluate the integral ∫(1/(x-2))^(2/3)dx from 1 to 4, we first perform a substitution. Let u = x - 2. Then du = dx. The integral becomes ∫(1/u)^(2/3)du. This simplifies to ∫u^(-2/3)du. Integrating this expression yields (3u^(1/3))/1/3 + C = 3u^(1/3) + C, where C is the constant of integration. Reversing the substitution, we get 3(x - 2)^(1/3) + C. Evaluating this expression from 1 to 4 gives (3(4 - 2)^(1/3)) - (3(1 - 2)^(1/3)) = 3(2^(1/3)) - 3(-1)^(1/3). Finally, simplifying this expression gives the value of the integral over the given interval.

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To evaluate the integral ( \int_{1}^{4} \frac{1}{{(x-2)^{2/3}}} , dx ), we can use the substitution method. Let ( u = x - 2 ). Then, ( du = dx ), and the limits of integration change accordingly.

When ( x = 1 ), ( u = 1 - 2 = -1 ). When ( x = 4 ), ( u = 4 - 2 = 2 ).

So, the integral becomes: [ \int_{-1}^{2} \frac{1}{{u^{2/3}}} , du ]

Now, we can rewrite the integrand as ( u^{-2/3} ), which makes it easier to integrate. Applying the power rule of integration: [ \int u^{-2/3} , du = \frac{{u^{1/3}}}{{1/3}} = 3u^{1/3} + C ]

Integrating from the limits ( -1 ) to ( 2 ): [ = \left[ 3u^{1/3} \right]_{-1}^{2} ] [ = 3(2^{1/3}) - 3(-1)^{1/3} ]

[ = 3(2^{1/3}) + 3 ]

Hence, the value of the integral ( \int_{1}^{4} \frac{1}{{(x-2)^{2/3}}} , dx ) is ( 3(2^{1/3}) + 3 ).

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