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

Lösung

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

\frac{1}{60}

Dezimale

0.01666 \dots

Schritte zur Lösung

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

Wende die partielle Integration an

= [\frac{1}{4} y^{4} (1 - y)^{2} - \int - \frac{1}{2} y^{4} (- y + 1) d y]_{0}^{1}

\int - \frac{1}{2} y^{4} (- y + 1) d y = - \frac{1}{2} (- \frac{y ^{6}}{6} + \frac{y ^{5}}{5})

= [\frac{1}{4} y^{4} (1 - y)^{2} - (- \frac{1}{2} (- \frac{y ^{6}}{6} + \frac{y ^{5}}{5}))]_{0}^{1}

Vereinfache

[\frac{1}{4} y^{4} (1 - y)^{2} - (- \frac{1}{2} (- \frac{y ^{6}}{6} + \frac{y ^{5}}{5}))]_{0}^{1} : [\frac{1}{60} (10 y^{6} - 24 y^{5} + 15 y^{4})]_{0}^{1}

= [\frac{1}{60} (10 y^{6} - 24 y^{5} + 15 y^{4})]_{0}^{1}

Berechne die Grenzen:

\frac{1}{60}

= \frac{1}{60}

\int_{1}^{4} x^{2} + 3 x + 2 d x

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