integral of (4x-x^2)^2

Lösung

\int (4 x - x^{2})^{2} d x

\frac{16 x ^{3}}{3} - 2 x^{4} + \frac{x ^{5}}{5} + C

Schritte zur Lösung

\int (4 x - x^{2})^{2} d x

Multipliziere aus

(4 x - x^{2})^{2} : 16 x^{2} - 8 x^{3} + x^{4}

= \int 16 x^{2} - 8 x^{3} + x^{4} d x

Wende die Summenregel an:

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

= \int 16 x^{2} d x - \int 8 x^{3} d x + \int x^{4} d x

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

\int 8 x^{3} d x = 2 x^{4}

\int x^{4} d x = \frac{x ^{5}}{5}

= \frac{16 x ^{3}}{3} - 2 x^{4} + \frac{x ^{5}}{5}

Füge eine Konstante zur Lösung hinzu

= \frac{16 x ^{3}}{3} - 2 x^{4} + \frac{x ^{5}}{5} + C

d er i v a t i v e e^{t} cos (2 t)

d er i v a t i v e y = e^{x + 1} + 1

d er i v a t i v e x^{2} + 7 x - 5

\int (\frac{3}{x} - \frac{x x}{4}) d x

z^{^{'}}