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

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

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

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

2 + \frac{π}{2}

Decimal

3.57079 \dots

Solution steps

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

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

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

= 2 + \frac{π}{2}

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