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

The integral from 0 to 4 of-x^2+2x+4sqrt(x) is 16

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

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

16

Solution steps

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

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

\int_{0}^{4} 2 x d x = 16

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

= - \frac{64}{3} + 16 + \frac{64}{3}

Simplify

= 16

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What is the integral from 0 to 4 of-x^2+2x+4sqrt(x) ?
The integral from 0 to 4 of-x^2+2x+4sqrt(x) is 16