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人気のある微分積分 >

integral of ysqrt(10+10y-25y^2)

解

\int y 10 + 10 y - 25 y^{2} d y

解

- \frac{1}{75} (- 25 y^{2} + 10 y + 10)^{\frac{3}{2}} + \frac{11}{50} (arcsin (\frac{1}{11} (5 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{5 y - 1}{11}))) + C

解答ステップ

\int y 10 + 10 y - 25 y^{2} d y

平方完成する

10 + 10 y - 25 y^{2} : - 25 (y - \frac{1}{5})^{2} + 11

= \int y - 25 (y - \frac{1}{5})^{2} + 11 d y

u置換積分法を適用する

= \int (u + \frac{1}{5}) - 25 u^{2} + 11 d u

拡張

(u + \frac{1}{5}) - 25 u^{2} + 11 : u - 25 u^{2} + 11 + \frac{- 25 u ^{2} + 11}{5}

= \int u - 25 u^{2} + 11 + \frac{- 25 u ^{2} + 11}{5} 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 u - 25 u^{2} + 11 d u + \int \frac{- 25 u ^{2} + 11}{5} d u

\int u - 25 u^{2} + 11 d u = - \frac{1}{75} (- 25 u^{2} + 11)^{\frac{3}{2}}

\int \frac{- 25 u ^{2} + 11}{5} d u = \frac{11}{50} (arcsin (\frac{5}{11} u) + \frac{1}{2} sin (2 arcsin (\frac{5}{11} u)))

= - \frac{1}{75} (- 25 u^{2} + 11)^{\frac{3}{2}} + \frac{11}{50} (arcsin (\frac{5}{11} u) + \frac{1}{2} sin (2 arcsin (\frac{5}{11} u)))

代用を戻す

u = y - \frac{1}{5}

= - \frac{1}{75} (- 25 (y - \frac{1}{5})^{2} + 11)^{\frac{3}{2}} + \frac{11}{50} (arcsin (\frac{5}{11} (y - \frac{1}{5})) + \frac{1}{2} sin (2 arcsin (\frac{5}{11} (y - \frac{1}{5}))))

簡素化

- \frac{1}{75} (- 25 (y - \frac{1}{5})^{2} + 11)^{\frac{3}{2}} + \frac{11}{50} (arcsin (\frac{5}{11} (y - \frac{1}{5})) + \frac{1}{2} sin (2 arcsin (\frac{5}{11} (y - \frac{1}{5})))) : - \frac{1}{75} (- 25 y^{2} + 10 y + 10)^{\frac{3}{2}} + \frac{11}{50} (arcsin (\frac{1}{11} (5 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{5 y - 1}{11})))

= - \frac{1}{75} (- 25 y^{2} + 10 y + 10)^{\frac{3}{2}} + \frac{11}{50} (arcsin (\frac{1}{11} (5 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{5 y - 1}{11})))

定数を解答に追加する

= - \frac{1}{75} (- 25 y^{2} + 10 y + 10)^{\frac{3}{2}} + \frac{11}{50} (arcsin (\frac{1}{11} (5 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{5 y - 1}{11}))) + C

グラフ

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