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

解

\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))))

解答ステップ

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

平方完成する

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

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

u置換積分法を適用する

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

拡張

(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

総和規則を適用する:

\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))))