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

The integral from 0 to 4 of (64sqrt(x)-8x^2) is 512/3

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integral from 0 to 4 of (64sqrt(x)-8x^2)

\int_{0}^{4} (64 x - 8 x^{2}) d x

\frac{512}{3}

Decimal Notation

170.66666 \dots

Solution steps

\int_{0}^{4} 64 x - 8 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} 64 x d x - \int_{0}^{4} 8 x^{2} d x

\int_{0}^{4} 64 x d x = \frac{1024}{3}

\int_{0}^{4} 8 x^{2} d x = \frac{512}{3}

= \frac{1024}{3} - \frac{512}{3}

Simplify

\frac{1024}{3} - \frac{512}{3} : \frac{512}{3}

= \frac{512}{3}

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What is the integral from 0 to 4 of (64sqrt(x)-8x^2) ?
The integral from 0 to 4 of (64sqrt(x)-8x^2) is 512/3