integral from 0 to 1-z-y of z

Lösung

\int_{0}^{1 - z - y} z d z

\frac{( - z + 1 - y ) ^{2}}{2}

Schritte zur Lösung

\int_{0}^{1 - z - y} z d z

Wende die Potenzregel an

= [\frac{z ^{2}}{2}]_{0}^{1 - y - z}

Berechne die Grenzen:

\frac{1}{2} (- z + 1 - y)^{2}

= \frac{1}{2} (- z + 1 - y)^{2}

Vereinfache

= \frac{( - z + 1 - y ) ^{2}}{2}

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