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

Lösung

\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))))

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Multipliziere aus

(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

Wende die Summenregel an:

\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))))

\int_{- 2}^{2} \frac{1}{2} (4 - x^{2})^{2} d x

\int_{1}^{e} (19 + ln (x))^{2} d x

\int_{1}^{4} \frac{1}{2 x + 3} d x

\int_{0}^{π} 2 sin^{2} (4 x) d x

\int_{0}^{\frac{π}{2}} \frac{1}{3 + 5 cos ( x )} d x