What is the integral from 2 to 4 of \sqrt[3]{x^5} ?

The integral from 2 to 4 of \sqrt[3]{x^5} is 6*4^{2/3}-3/(\sqrt[3]{2)}

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integral from 2 to 4 of \sqrt[3]{x^5}

\int_{2}^{4} 3 x^{5} d x

6 \cdot 4^{\frac{2}{3}} - \frac{3}{3 2}

Decimal

12.73795 \dots

Solution steps

\int_{2}^{4} 3 x^{5} d x

3 x^{5} = x^{\frac{5}{3}},

assuming

x \geq 0

= \int_{2}^{4} x^{\frac{5}{3}} d x

Apply the Power Rule

: [\frac{3}{8} x^{\frac{8}{3}}]_{2}^{4}

= [\frac{3}{8} x^{\frac{8}{3}}]_{2}^{4}

Compute the boundaries:

6 \cdot 4^{\frac{2}{3}} - \frac{3}{3 2}

= 6 \cdot 4^{\frac{2}{3}} - \frac{3}{3 2}

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What is the integral from 2 to 4 of \sqrt[3]{x^5} ?
The integral from 2 to 4 of \sqrt[3]{x^5} is 6*4^{2/3}-3/(\sqrt[3]{2)}