integral from 0 to 1 of 2pix(3x(x-2)^2)

Lösung

\int_{0}^{1} 2 π x (3 x (x - 2)^{2}) d x

\frac{16 π}{5}

Dezimale

10.05309 \dots

Schritte zur Lösung

\int_{0}^{1} 2 π x \cdot 3 x (x - 2)^{2} d x

Entferne die Konstante:

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

= 2 π 3 \cdot \int_{0}^{1} xx (x - 2)^{2} d x

Vereinfache

= 6 π \cdot \int_{0}^{1} xx (x - 2)^{2} d x

Multipliziere aus

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

= 6 π \cdot \int_{0}^{1} x^{4} - 4 x^{3} + 4 x^{2} 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

= 6 π (\int_{0}^{1} x^{4} d x - \int_{0}^{1} 4 x^{3} d x + \int_{0}^{1} 4 x^{2} d x)

\int_{0}^{1} x^{4} d x = \frac{1}{5}

\int_{0}^{1} 4 x^{3} d x = 1

\int_{0}^{1} 4 x^{2} d x = \frac{4}{3}

= 6 π (\frac{1}{5} - 1 + \frac{4}{3})

Vereinfache

6 π (\frac{1}{5} - 1 + \frac{4}{3}) : \frac{16 π}{5}

= \frac{16 π}{5}

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

\int_{- 2}^{- 1} \frac{1}{1 + x} d x

\int_{0}^{1} (3 x + 1) d x

\int_{0}^{2} 2^{4 x} d x

\int_{0}^{1} t 3^{t^{2}} d t