What is the integral from 0.5 to 1 of 6x(1-x) ?

The integral from 0.5 to 1 of 6x(1-x) is 0.5

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integral from 0.5 to 1 of 6x(1-x)

\int_{0.5}^{1} 6 x (1 - x) d x

0.5

Solution steps

\int_{0.5}^{1} 6 x (1 - x) d x

Take the constant out:

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

= 6 \cdot \int_{0.5}^{1} x (1 - x) d x

Expand

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

= 6 \cdot \int_{0.5}^{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

= 6 (\int_{0.5}^{1} x d x - \int_{0.5}^{1} x^{2} d x)

\int_{0.5}^{1} x d x = 0.375

\int_{0.5}^{1} x^{2} d x = 0.29166 \dots

= 6 (0.375 - 0.29166 \dots)

Simplify

= 0.5

What is the integral from 0.5 to 1 of 6x(1-x) ?
The integral from 0.5 to 1 of 6x(1-x) is 0.5