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人気のある微分積分 >

integral of ysqrt(6+4y-4y^2)

解

\int y 6 + 4 y - 4 y^{2} d y

解

- \frac{1}{12} (- 4 y^{2} + 4 y + 6)^{\frac{3}{2}} + \frac{7}{8} (arcsin (\frac{1}{7} (2 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 y - 1}{7}))) + C

解答ステップ

\int y 6 + 4 y - 4 y^{2} d y

平方完成する

6 + 4 y - 4 y^{2} : - 4 (y - \frac{1}{2})^{2} + 7

= \int y - 4 (y - \frac{1}{2})^{2} + 7 d y

u置換積分法を適用する

= \int (u + \frac{1}{2}) - 4 u^{2} + 7 d u

拡張

(u + \frac{1}{2}) - 4 u^{2} + 7 : u - 4 u^{2} + 7 + \frac{- 4 u ^{2} + 7}{2}

= \int u - 4 u^{2} + 7 + \frac{- 4 u ^{2} + 7}{2} 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 - 4 u^{2} + 7 d u + \int \frac{- 4 u ^{2} + 7}{2} d u

\int u - 4 u^{2} + 7 d u = - \frac{1}{12} (- 4 u^{2} + 7)^{\frac{3}{2}}

\int \frac{- 4 u ^{2} + 7}{2} d u = \frac{7}{8} (arcsin (\frac{2}{7} u) + \frac{1}{2} sin (2 arcsin (\frac{2}{7} u)))

= - \frac{1}{12} (- 4 u^{2} + 7)^{\frac{3}{2}} + \frac{7}{8} (arcsin (\frac{2}{7} u) + \frac{1}{2} sin (2 arcsin (\frac{2}{7} u)))

代用を戻す

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

= - \frac{1}{12} (- 4 (y - \frac{1}{2})^{2} + 7)^{\frac{3}{2}} + \frac{7}{8} (arcsin (\frac{2}{7} (y - \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{7} (y - \frac{1}{2}))))

簡素化

- \frac{1}{12} (- 4 (y - \frac{1}{2})^{2} + 7)^{\frac{3}{2}} + \frac{7}{8} (arcsin (\frac{2}{7} (y - \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{7} (y - \frac{1}{2})))) : - \frac{1}{12} (- 4 y^{2} + 4 y + 6)^{\frac{3}{2}} + \frac{7}{8} (arcsin (\frac{1}{7} (2 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 y - 1}{7})))

= - \frac{1}{12} (- 4 y^{2} + 4 y + 6)^{\frac{3}{2}} + \frac{7}{8} (arcsin (\frac{1}{7} (2 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 y - 1}{7})))

定数を解答に追加する

= - \frac{1}{12} (- 4 y^{2} + 4 y + 6)^{\frac{3}{2}} + \frac{7}{8} (arcsin (\frac{1}{7} (2 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 y - 1}{7}))) + C

グラフ

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