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

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

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

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

0

Solution steps

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

Expand

x (x - 2) (x - 4) : x^{3} - 6 x^{2} + 8 x

= \int_{0}^{4} x^{3} - 6 x^{2} + 8 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} 6 x^{2} d x + \int_{0}^{4} 8 x d x

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

\int_{0}^{4} 6 x^{2} d x = 128

\int_{0}^{4} 8 x d x = 64

= 64 - 128 + 64

Simplify

= 0

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