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

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

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

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

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

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

Take the constant out:

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

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

Expand

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

= 2 \cdot \int_{0}^{1} x - x^{2} 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

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

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

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

= 2 (\frac{1}{2} - \frac{1}{3})

Simplify

2 (\frac{1}{2} - \frac{1}{3}) : \frac{1}{3}

= \frac{1}{3}

2 tan^{3} (x)

\frac{\partial}{\partial x} (6 x y + 5 x^{2})

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

\int (sec (x) - cot (x))^{2} d x

\frac{x ^{9} + 7}{x ^{9}}

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