integral of pi(2x-4)^2

Lösung

\int π (2 x - 4)^{2} d x

π (\frac{4 x ^{3}}{3} - 8 x^{2} + 16 x) + C

Schritte zur Lösung

\int π (2 x - 4)^{2} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int (2 x - 4)^{2} d x

Multipliziere aus

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

= π \cdot \int 4 x^{2} - 16 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

= π (\int 4 x^{2} d x - \int 16 x d x + \int 16 d x)

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

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

\int 16 d x = 16 x

= π (\frac{4 x ^{3}}{3} - 8 x^{2} + 16 x)

Füge eine Konstante zur Lösung hinzu

= π (\frac{4 x ^{3}}{3} - 8 x^{2} + 16 x) + C

\frac{\partial}{\partial z} (3 x y)

\int_{0}^{7} 8 x \cdot 3 x + 1 d x

d er i v a t i v e f (x) = ln (x + 6)

\int_{0}^{15} 1000 e^{- 0.05 t} d t

x \to 1 lim (3 - 3 x)