integral from 4-y to 4-y^2 of y

Lösung

\int_{4 - y}^{4 - y^{2}} y d y

\frac{y ^{4} - 9 y ^{2} + 8 y}{2}

Schritte zur Lösung

\int_{4 - y}^{4 - y^{2}} y d y

Wende die Potenzregel an

= [\frac{y ^{2}}{2}]_{4 - y}^{4 - y^{2}}

Berechne die Grenzen:

\frac{( 4 - y ^{2} ) ^{2}}{2} - \frac{( 4 - y ) ^{2}}{2}

= \frac{( 4 - y ^{2} ) ^{2}}{2} - \frac{( 4 - y ) ^{2}}{2}

Vereinfache

= \frac{y ^{4} - 9 y ^{2} + 8 y}{2}

\int_{- y}^{2 y} x y - y^{2} d x

\int_{1}^{6} \frac{1}{1 + x ^{2}} d x

\int_{- 1}^{2} 2 x^{2} - 4 x + 5 d x

\int_{0}^{8} \frac{1}{v ^{\frac{1}{3}} + 1} d v

\int_{0}^{2} π (x^{2} - 4)^{2} d x