integral from a to b of (ax+b)^p

Lösung

\int_{a}^{b} (a x + b)^{p} d x

\frac{( b + ab ) ^{p + 1} - ( b + a ^{2} ) ^{p + 1}}{a ( p + 1 )}

Schritte zur Lösung

\int_{a}^{b} (a x + b)^{p} d x

Wende U-Substitution an

= \int_{b + a^{2}}^{b + ab} \frac{u ^{p}}{a} d u

Entferne die Konstante:

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

= \frac{1}{a} \cdot \int_{b + a^{2}}^{b + ab} u^{p} d u

Wende die Potenzregel an

= \frac{1}{a} [\frac{u ^{p + 1}}{p + 1}]_{b + a^{2}}^{b + ab}

Vereinfache

\frac{1}{a} [\frac{u ^{p + 1}}{p + 1}]_{b + a^{2}}^{b + ab} : \frac{[ \frac{u ^{p + 1}}{p + 1} ] _{b + a^{2}}^{b + ab}}{a}

= \frac{[ \frac{u ^{p + 1}}{p + 1} ] _{b + a^{2}}^{b + ab}}{a}

Berechne die Grenzen:

\frac{( b + ab ) ^{p + 1}}{p + 1} - \frac{( b + a ^{2} ) ^{p + 1}}{p + 1}

= \frac{\frac{( b + ab ) ^{p + 1}}{p + 1} - \frac{( b + a ^{2} ) ^{p + 1}}{p + 1}}{a}

Vereinfache

= \frac{( b + ab ) ^{p + 1} - ( b + a ^{2} ) ^{p + 1}}{a ( p + 1 )}

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