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

The integral from 1 to 9 of (9-x)^{1/3} is 12

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

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

12

Solution steps

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

Apply u-substitution

: \int_{8}^{0} - u^{\frac{1}{3}} d u

= \int_{8}^{0} - u^{\frac{1}{3}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{8} - u^{\frac{1}{3}} d u

Take the constant out:

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

= - (- \int_{0}^{8} u^{\frac{1}{3}} d u)

Apply the Power Rule

: [\frac{3}{4} u^{\frac{4}{3}}]_{0}^{8}

= - (- [\frac{3}{4} u^{\frac{4}{3}}]_{0}^{8})

Simplify

= [\frac{3}{4} u^{\frac{4}{3}}]_{0}^{8}

Compute the boundaries:

12

= 12

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What is the integral from 1 to 9 of (9-x)^{1/3} ?
The integral from 1 to 9 of (9-x)^{1/3} is 12