What is the integral from 0 to 2 of 6x(4-x) ?

The integral from 0 to 2 of 6x(4-x) is 32

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integral from 0 to 2 of 6x(4-x)

\int_{0}^{2} 6 x (4 - x) d x

32

Solution steps

\int_{0}^{2} 6 x (4 - 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}^{2} x (4 - x) d x

Expand

x (4 - x) : 4 x - x^{2}

= 6 \cdot \int_{0}^{2} 4 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}^{2} 4 x d x - \int_{0}^{2} x^{2} d x)

\int_{0}^{2} 4 x d x = 8

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

= 6 (8 - \frac{8}{3})

Simplify

6 (8 - \frac{8}{3}) : 32

= 32

What is the integral from 0 to 2 of 6x(4-x) ?
The integral from 0 to 2 of 6x(4-x) is 32