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Beliebt Rechnen >

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

Lösung

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

Lösung

- \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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Multipliziere aus

(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

Wende die Summenregel an:

\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)))

Setze in

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

ein

= - \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}))))

Vereinfache

- \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})))

Füge eine Konstante zur Lösung hinzu

= - \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

Graph

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