integral from 2y to 1+y^2 of z

Lösung

\int_{2 y}^{1 + y^{2}} z d z

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

Schritte zur Lösung

\int_{2 y}^{1 + y^{2}} z d z

Wende die Potenzregel an

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

Berechne die Grenzen:

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

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

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