What is the integral from-1 to 1 of (\sqrt[3]{t}-6) ?

The integral from-1 to 1 of (\sqrt[3]{t}-6) is -12

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integral from-1 to 1 of (\sqrt[3]{t}-6)

\int_{- 1}^{1} (3 t - 6) d t

- 12

Solution steps

\int_{- 1}^{1} 3 t - 6 d t

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 t d t - \int_{- 1}^{1} 6 d t

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

\int_{- 1}^{1} 6 d t = 12

= 0 - 12

Simplify

= - 12

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What is the integral from-1 to 1 of (\sqrt[3]{t}-6) ?
The integral from-1 to 1 of (\sqrt[3]{t}-6) is -12