What is the integral from-1 to 8 of \sqrt[3]{x^2} ?

The integral from-1 to 8 of \sqrt[3]{x^2} is 99/5

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integral from-1 to 8 of \sqrt[3]{x^2}

\frac{99}{5}

Decimal Notation

19.8

Solution steps

assuming

x \geq 0

= \int_{- 1}^{8} x^{\frac{2}{3}} d x

Apply the Power Rule

: [\frac{3}{5} x^{\frac{5}{3}}]_{- 1}^{8}

= [\frac{3}{5} x^{\frac{5}{3}}]_{- 1}^{8}

Compute the boundaries:

\frac{99}{5}

= \frac{99}{5}

\int_{0.4}^{1} x (1 - x)^{8} d x

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\int_{- 42.25}^{0} (\frac{13}{2} x)^{2} d x

\int_{0}^{x} \frac{1}{( 1 - x ) ^{\frac{2}{3}}} d x

What is the integral from-1 to 8 of \sqrt[3]{x^2} ?
The integral from-1 to 8 of \sqrt[3]{x^2} is 99/5