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

The integral from 0 to 1 of kx(1-x) is 1/6 k

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

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

\frac{1}{6} k

Solution steps

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

Take the constant out:

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

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

Expand

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

= k \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

= k (\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}

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

Join

\frac{1}{2} - \frac{1}{3} : \frac{1}{6}

= \frac{1}{6} k

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What is the integral from 0 to 1 of kx(1-x) ?
The integral from 0 to 1 of kx(1-x) is 1/6 k