What is the integral from-1 to 1 of (\sqrt[3]{t}-9) ?

The integral from-1 to 1 of (\sqrt[3]{t}-9) is -18

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integral from-1 to 1 of (\sqrt[3]{t}-9)

- 18

Solution steps

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} 9 d t = 18

= 0 - 18

Simplify

= - 18

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\int_{0}^{\frac{π}{2}} sin^{2} (2 t) d t

\int_{0}^{11} x d x

What is the integral from-1 to 1 of (\sqrt[3]{t}-9) ?
The integral from-1 to 1 of (\sqrt[3]{t}-9) is -18