What is the integral from 0 to 1 of (sqrt(2x-x^2)-x) ?

The integral from 0 to 1 of (sqrt(2x-x^2)-x) is pi/4-1/2

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integral from 0 to 1 of (sqrt(2x-x^2)-x)

\int_{0}^{1} (2 x - x^{2} - x) d x

\frac{π}{4} - \frac{1}{2}

Solution steps

\int_{0}^{1} 2 x - x^{2} - 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}^{1} 2 x - x^{2} d x - \int_{0}^{1} x d x

\int_{0}^{1} 2 x - x^{2} d x = \frac{π}{4}

\int_{0}^{1} x d x = \frac{1}{2}

= \frac{π}{4} - \frac{1}{2}

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What is the integral from 0 to 1 of (sqrt(2x-x^2)-x) ?
The integral from 0 to 1 of (sqrt(2x-x^2)-x) is pi/4-1/2