What is the integral from-1 to 1 of (\sqrt[3]{t}-2) ?

The integral from-1 to 1 of (\sqrt[3]{t}-2) is -4

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integral from-1 to 1 of (\sqrt[3]{t}-2)

- 4

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} 2 d t = 4

= 0 - 4

Simplify

= - 4

f (x) = \frac{4 x}{x ^{2} + 1}, at x = - 1

\int \frac{x}{x ^{2} + 3 x + 3} d x

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\frac{d}{d x} (x^{2} - 2 x)

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What is the integral from-1 to 1 of (\sqrt[3]{t}-2) ?
The integral from-1 to 1 of (\sqrt[3]{t}-2) is -4