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

The integral from 0 to 2 of kx(2-x) is 4/3 k

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

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

\frac{4}{3} k

Solution steps

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

Expand

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

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

\int_{0}^{2} 2 x d x = 4

\int_{0}^{2} x^{2} d x = \frac{8}{3}

= k (4 - \frac{8}{3})

Join

4 - \frac{8}{3} : \frac{4}{3}

= \frac{4}{3} k

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What is the integral from 0 to 2 of kx(2-x) ?
The integral from 0 to 2 of kx(2-x) is 4/3 k