What is the integral from-5 to 9 of \sqrt[3]{x} ?

The integral from-5 to 9 of \sqrt[3]{x} is (27\sqrt[3]{9}-15\sqrt[3]{5})/4

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integral from-5 to 9 of \sqrt[3]{x}

Decimal Notation

7.62815 \dots

Solution steps

Apply the Power Rule

: [\frac{3}{4} x^{\frac{4}{3}}]_{- 5}^{9}

= [\frac{3}{4} x^{\frac{4}{3}}]_{- 5}^{9}

Compute the boundaries:

\int_{- 3}^{0} 5 e^{- jω t} d t

\int_{0}^{\frac{π}{4}} sec^{4} (\frac{x}{4}) d x

\int_{0}^{\frac{x}{4}} e^{- x^{2}} d x

\int_{1}^{3} \frac{1}{x 4 + x ^{2}} d x

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

What is the integral from-5 to 9 of \sqrt[3]{x} ?
The integral from-5 to 9 of \sqrt[3]{x} is (27\sqrt[3]{9}-15\sqrt[3]{5})/4