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

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

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

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

\frac{1}{12}

Decimal Notation

0.08333 \dots

Solution steps

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

Expand

x (1 - x)^{2} : 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 t e^{3 t} d t

x \to + (- 2) lim (x^{2} - x + 4)

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 t e^{3 t}

f (x) = ln (9 sin^{2} (x))

\int \frac{3 x - 1}{x ^{2} - 4} d x

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