What is the integral from 1 to 9 of x\sqrt[3]{1-x} ?

The integral from 1 to 9 of x\sqrt[3]{1-x} is -468/7

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integral from 1 to 9 of x\sqrt[3]{1-x}

- \frac{468}{7}

Decimal Notation

- 66.85714 \dots

Solution steps

Apply u-substitution

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

Take the constant out:

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

Expand

Apply the Sum Rule:

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

Simplify

= - \frac{468}{7}

\int_{- 1}^{2} (2 x + 1)^{2} d x

\int_{2}^{3} 4 - x^{2} d x

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

\int_{1}^{2} \frac{1}{x + 1} d x

\int_{0}^{1} \frac{e ^{z} + 7}{e ^{z} + 7 z} d z

What is the integral from 1 to 9 of x\sqrt[3]{1-x} ?
The integral from 1 to 9 of x\sqrt[3]{1-x} is -468/7