integral of (x-4)^2

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

\int 16 d x = 16 x

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

Füge eine Konstante zur Lösung hinzu

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

x \to 0 + lim ((x + 1)^{I n x})

\frac{d v}{d t} = - k v

\frac{d y}{d x} - \frac{7 y}{x} = \frac{y ^{3}}{x ^{9}}

\frac{d}{d x} (\frac{3 x ^{2}}{2 x + 1})

\frac{d y}{d x} = \frac{( x - 2 )}{( y - 2 )}