What is the integral from 0 to 4 of (x(x-1)(x-2)) ?

The integral from 0 to 4 of (x(x-1)(x-2)) is 16

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integral from 0 to 4 of (x(x-1)(x-2))

\int_{0}^{4} (x (x - 1) (x - 2)) d x

16

Solution steps

\int_{0}^{4} x (x - 1) (x - 2) d x

Expand

x (x - 1) (x - 2) : x^{3} - 3 x^{2} + 2 x

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

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

\int_{0}^{4} x^{3} d x = 64

\int_{0}^{4} 3 x^{2} d x = 64

\int_{0}^{4} 2 x d x = 16

= 64 - 64 + 16

Simplify

= 16

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What is the integral from 0 to 4 of (x(x-1)(x-2)) ?
The integral from 0 to 4 of (x(x-1)(x-2)) is 16