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

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

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

\int_{0}^{1} 12 x (1 - x)^{2} d x

1

Solution steps

\int_{0}^{1} 12 x (1 - x)^{2} d x

Take the constant out:

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

= 12 \cdot \int_{0}^{1} x (1 - x)^{2} d x

Expand

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

= 12 \cdot \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

= 12 (\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}

= 12 (\frac{1}{2} - \frac{2}{3} + \frac{1}{4})

Simplify

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

= 1

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