What is the integral from 2 to 9 of 2x(x-1)^{1/3} ?

The integral from 2 to 9 of 2x(x-1)^{1/3} is 1839/14

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integral from 2 to 9 of 2x(x-1)^{1/3}

\int_{2}^{9} 2 x (x - 1)^{\frac{1}{3}} d x

\frac{1839}{14}

Decimal

131.35714 \dots

Solution steps

\int_{2}^{9} 2 x (x - 1)^{\frac{1}{3}} d x

Take the constant out:

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

= 2 \cdot \int_{2}^{9} x (x - 1)^{\frac{1}{3}} d x

Apply u-substitution

: \int_{1}^{2} 3 u^{3} (u^{3} + 1) d u

= 2 \cdot \int_{1}^{2} 3 u^{3} (u^{3} + 1) d u

Take the constant out:

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

= 2 \cdot 3 \cdot \int_{1}^{2} u^{3} (u^{3} + 1) d u

Apply Integration By Parts

: [\frac{1}{4} u^{4} (u^{3} + 1) - \int \frac{3 u ^{6}}{4} d u]_{1}^{2}

= 2 \cdot 3 [\frac{1}{4} u^{4} (u^{3} + 1) - \int \frac{3 u ^{6}}{4} d u]_{1}^{2}

\int \frac{3 u ^{6}}{4} d u = \frac{3 u ^{7}}{28}

= 2 \cdot 3 [\frac{1}{4} u^{4} (u^{3} + 1) - \frac{3 u ^{7}}{28}]_{1}^{2}

Simplify

2 \cdot 3 [\frac{1}{4} u^{4} (u^{3} + 1) - \frac{3 u ^{7}}{28}]_{1}^{2} : 6 [\frac{1}{28} (4 u^{7} + 7 u^{4})]_{1}^{2}

= 6 [\frac{1}{28} (4 u^{7} + 7 u^{4})]_{1}^{2}

Compute the boundaries:

\frac{613}{28}

= 6 \cdot \frac{613}{28}

Simplify

= \frac{1839}{14}

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What is the integral from 2 to 9 of 2x(x-1)^{1/3} ?
The integral from 2 to 9 of 2x(x-1)^{1/3} is 1839/14