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}

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

\frac{- 3 3 - 1 + 9 3 - 1 3 3}{4}

Solution steps

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

Apply u-substitution

: \int_{- 3}^{- 1} 3 u d u

= \int_{- 3}^{- 1} 3 u d u

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:

\frac{- 3 3 - 1 + 9 3 - 1 3 3}{4}

= \frac{- 3 3 - 1 + 9 3 - 1 3 3}{4}

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