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 (-3\sqrt[3]{-1}+9\sqrt[3]{-1}\sqrt[3]{3})/4

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

Solution steps

Apply u-substitution

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{- 3}^{- 1}

= [\frac{3}{4} u^{\frac{4}{3}}]_{- 3}^{- 1}

Compute the boundaries:

y = 1.1 x + e^{x}, (0, 1)

y = 0, y = x^{2}, y = 4 x - x^{2}, [0, 2]

(1 + cos (t)) y^{^{'}} = (1 + e^{- y}) sin (t), y (0) = 0

(x^{2} + 5) (8 - x)

\int \frac{2 x}{1 + 4 x ^{2}} d x

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 (-3\sqrt[3]{-1}+9\sqrt[3]{-1}\sqrt[3]{3})/4