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

The integral from 4 to 0 of 4x-x^2 is -32/3

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integral from 4 to 0 of 4x-x^2

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

- \frac{32}{3}

Decimal

- 10.66666 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{4} 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

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

\int_{0}^{4} 4 x d x = 32

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

= - (32 - \frac{64}{3})

Join

32 - \frac{64}{3} : \frac{32}{3}

= - \frac{32}{3}

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What is the integral from 4 to 0 of 4x-x^2 ?
The integral from 4 to 0 of 4x-x^2 is -32/3