What is the integral from 0 to 3 of (2x-x^2)^4(1-x) ?

The integral from 0 to 3 of (2x-x^2)^4(1-x) is -243/10

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integral from 0 to 3 of (2x-x^2)^4(1-x)

\int_{0}^{3} (2 x - x^{2})^{4} (1 - x) d x

- \frac{243}{10}

Decimal Notation

- 24.3

Solution steps

\int_{0}^{3} (2 x - x^{2})^{4} (1 - x) d x

Apply u-substitution

: \int_{0}^{- 3} \frac{u ^{4}}{2} d u

= \int_{0}^{- 3} \frac{u ^{4}}{2} d u

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

= - \int_{- 3}^{0} \frac{u ^{4}}{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^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{- 3}^{0}

= - \frac{1}{2} [\frac{u ^{5}}{5}]_{- 3}^{0}

Compute the boundaries:

\frac{243}{5}

= - \frac{1}{2} \cdot \frac{243}{5}

Simplify

= - \frac{243}{10}

\frac{d}{d x} (\frac{1}{4} (x^{2} - 4))

\frac{\partial}{\partial x} (2 x y - 18 x)

\frac{3 x}{x ^{2} + 1}

\frac{d}{d x} (- x^{3} + 2 x + 7)

- \frac{5}{s ^{2} + 2 s + 4}

What is the integral from 0 to 3 of (2x-x^2)^4(1-x) ?
The integral from 0 to 3 of (2x-x^2)^4(1-x) is -243/10