integral from 0 to 1 of (2-y)^2sqrt(y)

Lösung

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

\frac{142}{105}

Dezimale

1.35238 \dots

Schritte zur Lösung

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

Multipliziere aus

(2 - y)^{2} y : 4 y - 4 y^{\frac{3}{2}} + y^{\frac{5}{2}}

= \int_{0}^{1} 4 y - 4 y^{\frac{3}{2}} + y^{\frac{5}{2}} d y

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}^{1} 4 y d y - \int_{0}^{1} 4 y^{\frac{3}{2}} d y + \int_{0}^{1} y^{\frac{5}{2}} d y

\int_{0}^{1} 4 y d y = \frac{8}{3}

\int_{0}^{1} 4 y^{\frac{3}{2}} d y = \frac{8}{5}

\int_{0}^{1} y^{\frac{5}{2}} d y = \frac{2}{7}

= \frac{8}{3} - \frac{8}{5} + \frac{2}{7}

Vereinfache

\frac{8}{3} - \frac{8}{5} + \frac{2}{7} : \frac{142}{105}

= \frac{142}{105}