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

The integral from 0 to 2 of 3x(x-4) is -16

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integral from 0 to 2 of 3x(x-4)

\int_{0}^{2} 3 x (x - 4) d x

- 16

Solution steps

\int_{0}^{2} 3 x (x - 4) d x

Take the constant out:

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

= 3 \cdot \int_{0}^{2} x (x - 4) d x

Expand

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

= 3 \cdot \int_{0}^{2} x^{2} - 4 x 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

= 3 (\int_{0}^{2} x^{2} d x - \int_{0}^{2} 4 x d x)

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

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

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

Simplify

3 (\frac{8}{3} - 8) : - 16

= - 16

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What is the integral from 0 to 2 of 3x(x-4) ?
The integral from 0 to 2 of 3x(x-4) is -16