What is the integral from 0 to 8 of (8x-x^2) ?

The integral from 0 to 8 of (8x-x^2) is 256/3

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integral from 0 to 8 of (8x-x^2)

\int_{0}^{8} (8 x - x^{2}) d x

\frac{256}{3}

Decimal Notation

85.33333 \dots

Solution steps

\int_{0}^{8} 8 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}^{8} 8 x d x - \int_{0}^{8} x^{2} d x

\int_{0}^{8} 8 x d x = 256

\int_{0}^{8} x^{2} d x = \frac{512}{3}

= 256 - \frac{512}{3}

Simplify

256 - \frac{512}{3} : \frac{256}{3}

= \frac{256}{3}

\int (8 x - 12) (4 x^{2} - 12 x)^{4} d x

\frac{d}{d y} (e^{2 x + y})

f (x) = arctan (x)

\frac{d}{d x} (x^{2} \cdot e^{- 4 x})

\frac{d}{d x} ((x - 3) x)

What is the integral from 0 to 8 of (8x-x^2) ?
The integral from 0 to 8 of (8x-x^2) is 256/3