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

The integral from-1 to 1 of (\sqrt[3]{x}-8) is -16

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integral from-1 to 1 of (\sqrt[3]{x}-8)

\int_{- 1}^{1} (3 x - 8) d x

- 16

Solution steps

\int_{- 1}^{1} 3 x - 8 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{- 1}^{1} 3 x d x = 0

\int_{- 1}^{1} 8 d x = 16

= 0 - 16

Simplify

= - 16

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What is the integral from-1 to 1 of (\sqrt[3]{x}-8) ?
The integral from-1 to 1 of (\sqrt[3]{x}-8) is -16