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

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

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integral from 1 to k of \sqrt[3]{x-1}

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

\frac{3 ( k - 1 ) ^{\frac{4}{3}}}{4}

Solution steps

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

Apply u-substitution

: \int_{0}^{k - 1} 3 u d u

= \int_{0}^{k - 1} 3 u d u

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{0}^{k - 1}

= [\frac{3}{4} u^{\frac{4}{3}}]_{0}^{k - 1}

Compute the boundaries:

\frac{3}{4} (k - 1)^{\frac{4}{3}}

= \frac{3}{4} (k - 1)^{\frac{4}{3}}

Simplify

= \frac{3 ( k - 1 ) ^{\frac{4}{3}}}{4}

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