What is the integral from 0 to 1 of x^2\sqrt[3]{1-x} ?

The integral from 0 to 1 of x^2\sqrt[3]{1-x} is 27/140

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integral from 0 to 1 of x^2\sqrt[3]{1-x}

\frac{27}{140}

Decimal Notation

0.19285 \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

\int_{0}^{1} u^{\frac{7}{3}} d u = \frac{3}{10}

\int_{0}^{1} 2 u^{\frac{4}{3}} d u = \frac{6}{7}

= - (- (\frac{3}{10} - \frac{6}{7} + \frac{3}{4}))

Simplify

- (- (\frac{3}{10} - \frac{6}{7} + \frac{3}{4})) : \frac{27}{140}

= \frac{27}{140}

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What is the integral from 0 to 1 of x^2\sqrt[3]{1-x} ?
The integral from 0 to 1 of x^2\sqrt[3]{1-x} is 27/140