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受欢迎的微积分 >

integral of ysqrt(4+4y-4y^2)

解答

\int y 4 + 4 y - 4 y^{2} d y

解答

- \frac{2}{3} (- y^{2} + y + 1)^{\frac{3}{2}} + \frac{5}{8} (arcsin (\frac{1}{5} (2 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 y - 1}{5}))) + C

求解步骤

\int y 4 + 4 y - 4 y^{2} d y

对

4 + 4 y - 4 y^{2}

配方:

- 4 (y - \frac{1}{2})^{2} + 5

= \int y - 4 (y - \frac{1}{2})^{2} + 5 d y

使用换元积分法

= \int (u + \frac{1}{2}) - 4 u^{2} + 5 d u

乘开

(u + \frac{1}{2}) - 4 u^{2} + 5 : u - 4 u^{2} + 5 + \frac{- 4 u ^{2} + 5}{2}

= \int u - 4 u^{2} + 5 + \frac{- 4 u ^{2} + 5}{2} d u

使用积分加法定则:

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

= \int u - 4 u^{2} + 5 d u + \int \frac{- 4 u ^{2} + 5}{2} d u

\int u - 4 u^{2} + 5 d u = - \frac{1}{12} (- 4 u^{2} + 5)^{\frac{3}{2}}

\int \frac{- 4 u ^{2} + 5}{2} d u = \frac{5}{8} (arcsin (\frac{2}{5} u) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} u)))

= - \frac{1}{12} (- 4 u^{2} + 5)^{\frac{3}{2}} + \frac{5}{8} (arcsin (\frac{2}{5} u) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} u)))

u = y - \frac{1}{2}

代回

= - \frac{1}{12} (- 4 (y - \frac{1}{2})^{2} + 5)^{\frac{3}{2}} + \frac{5}{8} (arcsin (\frac{2}{5} (y - \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (y - \frac{1}{2}))))

化简

- \frac{1}{12} (- 4 (y - \frac{1}{2})^{2} + 5)^{\frac{3}{2}} + \frac{5}{8} (arcsin (\frac{2}{5} (y - \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (y - \frac{1}{2})))) : - \frac{2}{3} (- y^{2} + y + 1)^{\frac{3}{2}} + \frac{5}{8} (arcsin (\frac{1}{5} (2 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 y - 1}{5})))

= - \frac{2}{3} (- y^{2} + y + 1)^{\frac{3}{2}} + \frac{5}{8} (arcsin (\frac{1}{5} (2 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 y - 1}{5})))

解答补常数

= - \frac{2}{3} (- y^{2} + y + 1)^{\frac{3}{2}} + \frac{5}{8} (arcsin (\frac{1}{5} (2 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 y - 1}{5}))) + C

作图

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