integral from 0 to 1 of v(u+v^2)^4

Lösung

\int_{0}^{1} v (u + v^{2})^{4} d u

\frac{v ( 5 v ^{8} + 10 v ^{6} + 10 v ^{4} + 5 v ^{2} + 1 )}{5}

Schritte zur Lösung

\int_{0}^{1} v (u + v^{2})^{4} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= v \cdot \int_{0}^{1} (v^{2} + u)^{4} d u

Wende U-Substitution an

= v \cdot \int_{v^{2}}^{v^{2} + 1} w^{4} d w

Wende die Potenzregel an

= v [\frac{w ^{5}}{5}]_{v^{2}}^{v^{2} + 1}

Berechne die Grenzen:

\frac{( v ^{2} + 1 ) ^{5}}{5} - \frac{v ^{10}}{5}

= v (\frac{( v ^{2} + 1 ) ^{5}}{5} - \frac{v ^{10}}{5})

Vereinfache

= \frac{v ( 5 v ^{8} + 10 v ^{6} + 10 v ^{4} + 5 v ^{2} + 1 )}{5}