How do you evaluate the integral $\int \frac{1}{\sqrt[3]{x}}$ $int 1/root3x$ from -1 to 1?

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How do you evaluate the integral $\int \frac{1}{\sqrt[3]{x}}$ $int 1/root3x$ from -1 to 1?

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Answer 1 · 2016-09-29T04:13:19

$\int_{- 1}^{1} x^{- \frac{1}{3}} = 0$ $int_-1^1x^(-1/3) = 0$

Explanation:

Expressing the reciprocal third root of x as $x^{- \frac{1}{3}}$ $x^(-1/3)$ is useful here, as we can simply apply the reverse power rule. It is also important to note that there is a discontinuity at $x = 0$ $x=0$ , so we must split the integral into two parts:

$\int_{0}^{1} x^{- \frac{1}{3}} + \int_{- 1}^{0} x^{- \frac{1}{3}}$ $int_0^1x^(-1/3) + int_-1^0x^(-1/3)$

By the reverse power rule, we increment the power by one and divide by the exponent:

$\frac{3}{2} x^{\frac{2}{3}} ∣_{0}^{1} = \frac{3}{2} - 0$ $3/2x^(2/3)|_0^1 = 3/2 - 0$

$+$ $+$ $\frac{3}{2} x^{\frac{2}{3}} ∣_{- 1}^{0} = 0 - \frac{3}{2}$ $3/2x^(2/3)|_-1^0 = 0 -3/2$

Thus,
$\int_{- 1}^{1} x^{- \frac{1}{3}} = \frac{3}{2} - \frac{3}{2} = 0$ $int_-1^1x^(-1/3) = 3/2 -3/2 = 0$

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