What is the integral from 0 to 3 of kx(3-x) ?

The integral from 0 to 3 of kx(3-x) is 9/2 k

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integral from 0 to 3 of kx(3-x)

\int_{0}^{3} k x (3 - x) d x

\frac{9}{2} k

Solution steps

\int_{0}^{3} k x (3 - 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}^{3} x (3 - x) d x

Expand

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

= k \cdot \int_{0}^{3} 3 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}^{3} 3 x d x - \int_{0}^{3} x^{2} d x)

\int_{0}^{3} 3 x d x = \frac{27}{2}

\int_{0}^{3} x^{2} d x = 9

= k (\frac{27}{2} - 9)

Join

\frac{27}{2} - 9 : \frac{9}{2}

= \frac{9}{2} k

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What is the integral from 0 to 3 of kx(3-x) ?
The integral from 0 to 3 of kx(3-x) is 9/2 k