What is the integral from-7 to 28 of x\sqrt[3]{1-x} ?

The integral from-7 to 28 of x\sqrt[3]{1-x} is -29145/28

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integral from-7 to 28 of x\sqrt[3]{1-x}

\int_{- 7}^{28} x 3 1 - x d x

- \frac{29145}{28}

Decimal

- 1040.89285 \dots

Solution steps

\int_{- 7}^{28} x 3 1 - x d x

Apply u-substitution

: \int_{8}^{- 27} - (- u + 1) 3 u d u

= \int_{8}^{- 27} - (- u + 1) 3 u d u

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

= - \int_{- 27}^{8} - (- u + 1) 3 u d u

Take the constant out:

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

= - (- \int_{- 27}^{8} (- u + 1) 3 u d u)

Expand

(- u + 1) 3 u : - u^{\frac{4}{3}} + 3 u

= - (- \int_{- 27}^{8} - u^{\frac{4}{3}} + 3 u d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (- \int_{- 27}^{8} u^{\frac{4}{3}} d u + \int_{- 27}^{8} 3 u d u))

\int_{- 27}^{8} u^{\frac{4}{3}} d u = \frac{6945}{7}

\int_{- 27}^{8} 3 u d u = - \frac{195}{4}

= - (- (- \frac{6945}{7} - \frac{195}{4}))

Simplify

- (- (- \frac{6945}{7} - \frac{195}{4})) : - \frac{29145}{28}

= - \frac{29145}{28}

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What is the integral from-7 to 28 of x\sqrt[3]{1-x} ?
The integral from-7 to 28 of x\sqrt[3]{1-x} is -29145/28