What is the integral from-2 to 0 of \sqrt[3]{x^2} ?

The integral from-2 to 0 of \sqrt[3]{x^2} is (6(-1)^{2/3}*2^{2/3})/5

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integral from-2 to 0 of \sqrt[3]{x^2}

\frac{6 ( - 1 ) ^{\frac{2}{3}} \cdot 2 ^{\frac{2}{3}}}{5}

Solution steps

assuming

x \geq 0

= \int_{- 2}^{0} x^{\frac{2}{3}} d x

Apply the Power Rule

: [\frac{3}{5} x^{\frac{5}{3}}]_{- 2}^{0}

= [\frac{3}{5} x^{\frac{5}{3}}]_{- 2}^{0}

Compute the boundaries:

\frac{6 ( - 1 ) ^{\frac{2}{3}} \cdot 2 ^{\frac{2}{3}}}{5}

= \frac{6 ( - 1 ) ^{\frac{2}{3}} \cdot 2 ^{\frac{2}{3}}}{5}

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What is the integral from-2 to 0 of \sqrt[3]{x^2} ?
The integral from-2 to 0 of \sqrt[3]{x^2} is (6(-1)^{2/3}*2^{2/3})/5