integral from 0 to 1 of y(-3y^2+2y+1)

Lösung

\int_{0}^{1} y (- 3 y^{2} + 2 y + 1) d y

\frac{5}{12}

Dezimale

0.41666 \dots

Schritte zur Lösung

\int_{0}^{1} y (- 3 y^{2} + 2 y + 1) d y

Multipliziere aus

y (- 3 y^{2} + 2 y + 1) : - 3 y^{3} + 2 y^{2} + y

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

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

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

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

= - \frac{3}{4} + \frac{2}{3} + \frac{1}{2}

Vereinfache

- \frac{3}{4} + \frac{2}{3} + \frac{1}{2} : \frac{5}{12}

= \frac{5}{12}

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