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

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

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

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

- 4

Solution steps

\int_{- 1}^{1} 3 x - 2 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} 2 d x

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

\int_{- 1}^{1} 2 d x = 4

= 0 - 4

Simplify

= - 4

t an g e n t y = \frac{3}{x}

\int sin^{2} (x) cos^{2} (2 x) d x

\frac{d}{d x} (f (x) x)

x \to - 3 + lim (\frac{∣ x + 3 ∣}{x + 3})

\int_{- 4}^{0} y + 4 d y

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