What is the integral from 0 to 3 of (2x-x^2)^3(1-x) ?

The integral from 0 to 3 of (2x-x^2)^3(1-x) is 81/8

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integral from 0 to 3 of (2x-x^2)^3(1-x)

\int_{0}^{3} (2 x - x^{2})^{3} (1 - x) d x

\frac{81}{8}

Decimal Notation

10.125

Solution steps

\int_{0}^{3} (2 x - x^{2})^{3} (1 - x) d x

Apply u-substitution

: \int_{0}^{- 3} \frac{u ^{3}}{2} d u

= \int_{0}^{- 3} \frac{u ^{3}}{2} d u

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

= - \int_{- 3}^{0} \frac{u ^{3}}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int_{- 3}^{0} u^{3} d u

Apply the Power Rule

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

= - \frac{1}{2} [\frac{u ^{4}}{4}]_{- 3}^{0}

Compute the boundaries:

- \frac{81}{4}

= - \frac{1}{2} (- \frac{81}{4})

Simplify

= \frac{81}{8}

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What is the integral from 0 to 3 of (2x-x^2)^3(1-x) ?
The integral from 0 to 3 of (2x-x^2)^3(1-x) is 81/8