integral from 0 to H of z^2(1-z)^2

Lösung

\int_{0}^{H} z^{2} (1 - z)^{2} d z

\frac{H ^{3}}{3} - \frac{H ^{4}}{2} + \frac{H ^{5}}{5}

Schritte zur Lösung

\int_{0}^{H} z^{2} (1 - z)^{2} d z

Multipliziere aus

z^{2} (1 - z)^{2} : z^{2} - 2 z^{3} + z^{4}

= \int_{0}^{H} z^{2} - 2 z^{3} + z^{4} d z

Wende die Summenregel an:

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

= \int_{0}^{H} z^{2} d z - \int_{0}^{H} 2 z^{3} d z + \int_{0}^{H} z^{4} d z

\int_{0}^{H} z^{2} d z = \frac{H ^{3}}{3}

\int_{0}^{H} 2 z^{3} d z = \frac{H ^{4}}{2}

\int_{0}^{H} z^{4} d z = \frac{H ^{5}}{5}

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

(x^{2} + y) + (x + e^{y}) y^{^{'}} = 0

\frac{d y}{d x} = \frac{cos ( 10 x )}{e ^{10 y}}

x \to 0 lim (\frac{2 ^{x}}{3 ^{x}})

x \to 0 lim (\frac{x ^{2}}{x ^{2} + 1})

\int_{1}^{2} \frac{9}{x ^{3} + 5 x} d x