# How do you find the integral of #int (dx / x^(2/3)) # from 1 to -15?

So for the first integral we get:

And for the second integral we get:

Both improper integrals converge, so we get:

Note on limits of integration

then the answer should be

And the details of the solution should be changed appropriately.

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To find the integral of ( \int_{1}^{-15} \frac{dx}{x^{2/3}} ), we can integrate the function over the given interval using the definite integral formula:

[ \int_{1}^{-15} \frac{dx}{x^{2/3}} ]

First, simplify the integral by rewriting ( x^{-2/3} ) as ( \frac{1}{x^{2/3}} ). Then integrate:

[ \int_{1}^{-15} x^{-2/3} , dx = \int_{1}^{-15} \frac{1}{x^{2/3}} , dx ]

The integral of ( x^{-2/3} ) is ( \frac{3x^{1/3}}{1/3} + C ), where ( C ) is the constant of integration. Simplify this to ( 9x^{1/3} + C ).

Now plug in the upper and lower limits:

[ \left[ 9x^{1/3} \right]_{1}^{-15} ]

[ = 9(-15)^{1/3} - 9(1)^{1/3} ]

[ = 9(-15^{1/3}) - 9(1) ]

[ = -27 ]

Therefore, the integral of ( \int_{1}^{-15} \frac{dx}{x^{2/3}} ) from 1 to -15 is -27.

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