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

The integral from 0 to 1 of (1-x)^2x is 1/12

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

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

\frac{1}{12}

Decimal Notation

0.08333 \dots

Solution steps

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

Expand

(1 - x)^{2} x : x - 2 x^{2} + x^{3}

= \int_{0}^{1} x - 2 x^{2} + x^{3} d x

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

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 2 x^{2} d x = \frac{2}{3}

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= \frac{1}{2} - \frac{2}{3} + \frac{1}{4}

Simplify

\frac{1}{2} - \frac{2}{3} + \frac{1}{4} : \frac{1}{12}

= \frac{1}{12}

\int (3 x - \frac{2}{x}) d x

\frac{d y}{d x} = 20 x + y - 20

\int_{7}^{\infty} \frac{9}{v ^{2} - v} d v

\frac{\partial}{\partial x} (e^{\frac{x + y}{x - y}})

(10.2) (19.14)

What is the integral from 0 to 1 of (1-x)^2x ?
The integral from 0 to 1 of (1-x)^2x is 1/12