What is the integral from 0 to 8 of \sqrt[3]{x} ?

The integral from 0 to 8 of \sqrt[3]{x} is 12

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integral from 0 to 8 of \sqrt[3]{x}

\int_{0}^{8} 3 x d x

12

Solution steps

\int_{0}^{8} 3 x d x

Apply the Power Rule

: [\frac{3}{4} x^{\frac{4}{3}}]_{0}^{8}

= [\frac{3}{4} x^{\frac{4}{3}}]_{0}^{8}

Compute the boundaries:

12

= 12

\frac{d y}{d x} + 4 y^{2} = 0

\int \frac{1}{t t ^{2} - 1} d t

f (x) = \frac{6}{x}, at x = 5

\int_{0}^{4} t e^{- t} d t

\frac{400}{25 s ( s ^{2} + 200 )}

What is the integral from 0 to 8 of \sqrt[3]{x} ?
The integral from 0 to 8 of \sqrt[3]{x} is 12