What is the integral from 0 to 8 of sqrt(2x)-(x-4) ?

The integral from 0 to 8 of sqrt(2x)-(x-4) is 64/3

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integral from 0 to 8 of sqrt(2x)-(x-4)

\int_{0}^{8} 2 x - (x - 4) d x

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

\int_{0}^{8} 2 x - (x - 4) d x

Simplify

2 x - (x - 4) : 2 x - x + 4

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

\int_{0}^{8} 2 x d x = \frac{64}{3}

\int_{0}^{8} x d x = 32

\int_{0}^{8} 4 d x = 32

= \frac{64}{3} - 32 + 32

Simplify

= \frac{64}{3}

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What is the integral from 0 to 8 of sqrt(2x)-(x-4) ?
The integral from 0 to 8 of sqrt(2x)-(x-4) is 64/3