integral from 0 to 2 of 2pix(2x^2-x^3)

Lösung

\int_{0}^{2} 2 π x (2 x^{2} - x^{3}) d x

\frac{16 π}{5}

Dezimale

10.05309 \dots

Schritte zur Lösung

\int_{0}^{2} 2 π x (2 x^{2} - x^{3}) d x

Entferne die Konstante:

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

= 2 π \cdot \int_{0}^{2} x (2 x^{2} - x^{3}) d x

Multipliziere aus

x (2 x^{2} - x^{3}) : 2 x^{3} - x^{4}

= 2 π \cdot \int_{0}^{2} 2 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

= 2 π (\int_{0}^{2} 2 x^{3} d x - \int_{0}^{2} x^{4} d x)

\int_{0}^{2} 2 x^{3} d x = 8

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= 2 π (8 - \frac{32}{5})

Vereinfache

2 π (8 - \frac{32}{5}) : \frac{16 π}{5}

= \frac{16 π}{5}

\int \frac{1}{( a + b ) - ( a - b ) x ^{2}} d x

\int \frac{22}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (2 x arcsin (\frac{x}{y}))

a re a y = x^{- 3}, 1 \leq x \leq 5

d er i v a t i v e y = \frac{- 1}{3 sin ^{3} ( x )}