What is the integral from 1 to 5 of \sqrt[3]{2x-1} ?

The integral from 1 to 5 of \sqrt[3]{2x-1} is (27\sqrt[3]{9}-3)/8

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integral from 1 to 5 of \sqrt[3]{2x-1}

\int_{1}^{5} 3 2 x - 1 d x

\frac{27 3 9 - 3}{8}

Decimal

6.64528 \dots

Solution steps

\int_{1}^{5} 3 2 x - 1 d x

Apply u-substitution

: \int_{1}^{9} \frac{3 u}{2} d u

= \int_{1}^{9} \frac{3 u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1}^{9} 3 u d u

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{1}^{9}

= \frac{1}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{1}^{9}

Compute the boundaries:

\frac{27 3 9}{4} - \frac{3}{4}

= \frac{1}{2} (\frac{27 3 9}{4} - \frac{3}{4})

Simplify

= \frac{27 3 9 - 3}{8}

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x \to - 4 + lim (\frac{∣ x + 4 ∣}{x + 4})

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What is the integral from 1 to 5 of \sqrt[3]{2x-1} ?
The integral from 1 to 5 of \sqrt[3]{2x-1} is (27\sqrt[3]{9}-3)/8