What is the integral from a to b of xsqrt(2x-x^2) ?

The integral from a to b of xsqrt(2x-x^2) is

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integral from a to b of xsqrt(2x-x^2)

\int_{a}^{b} x 2 x - x^{2} d x

- \frac{( - b ^{2} + 2 b ) ^{\frac{3}{2}} - ( - a ^{2} + 2 a ) ^{\frac{3}{2}}}{3} + \frac{1}{2} (arcsin (b - 1) - arcsin (a - 1) + \frac{1}{2} (sin (2 arcsin (b - 1)) - sin (2 arcsin (a - 1))))

Solution steps

\int_{a}^{b} x 2 x - x^{2} d x

Complete the square

2 x - x^{2} : - (x - 1)^{2} + 1

= \int_{a}^{b} x - (x - 1)^{2} + 1 d x

Apply u-substitution

: \int_{a - 1}^{b - 1} (u + 1) - u^{2} + 1 d u

= \int_{a - 1}^{b - 1} (u + 1) - u^{2} + 1 d u

Expand

(u + 1) - u^{2} + 1 : u - u^{2} + 1 + - u^{2} + 1

= \int_{a - 1}^{b - 1} u - u^{2} + 1 + - u^{2} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{a - 1}^{b - 1} u - u^{2} + 1 d u + \int_{a - 1}^{b - 1} - u^{2} + 1 d u

\int_{a - 1}^{b - 1} u - u^{2} + 1 d u = - \frac{( - b ^{2} + 2 b ) ^{\frac{3}{2}} - ( - a ^{2} + 2 a ) ^{\frac{3}{2}}}{3}

\int_{a - 1}^{b - 1} - u^{2} + 1 d u = \frac{1}{2} (arcsin (b - 1) - arcsin (a - 1) + \frac{1}{2} (sin (2 arcsin (b - 1)) - sin (2 arcsin (a - 1))))

= - \frac{( - b ^{2} + 2 b ) ^{\frac{3}{2}} - ( - a ^{2} + 2 a ) ^{\frac{3}{2}}}{3} + \frac{1}{2} (arcsin (b - 1) - arcsin (a - 1) + \frac{1}{2} (sin (2 arcsin (b - 1)) - sin (2 arcsin (a - 1))))

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What is the integral from a to b of xsqrt(2x-x^2) ?
The integral from a to b of xsqrt(2x-x^2) is