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

The integral from 0 to 6 of (2x)-(x^2-4x) is 36

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

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

36

Solution steps

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

Simplify

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

\int_{0}^{6} x^{2} d x = 72

\int_{0}^{6} 6 x d x = 108

= - 72 + 108

Simplify

= 36

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What is the integral from 0 to 6 of (2x)-(x^2-4x) ?
The integral from 0 to 6 of (2x)-(x^2-4x) is 36