integral from a to b of (Cx+D)^p

Lösung

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

\frac{( D + C b ) ^{p + 1} - ( D + C a ) ^{p + 1}}{C ( p + 1 )}

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{D + C a}^{D + C b} \frac{u ^{p}}{C} d u

Entferne die Konstante:

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

= \frac{1}{C} \cdot \int_{D + C a}^{D + C b} u^{p} d u

Wende die Potenzregel an

= \frac{1}{C} [\frac{u ^{p + 1}}{p + 1}]_{D + C a}^{D + C b}

Vereinfache

\frac{1}{C} [\frac{u ^{p + 1}}{p + 1}]_{D + C a}^{D + C b} : \frac{[ \frac{u ^{p + 1}}{p + 1} ] _{D + C a}^{D + C b}}{C}

= \frac{[ \frac{u ^{p + 1}}{p + 1} ] _{D + C a}^{D + C b}}{C}

Berechne die Grenzen:

\frac{( D + C b ) ^{p + 1}}{p + 1} - \frac{( D + C a ) ^{p + 1}}{p + 1}

= \frac{\frac{( D + C b ) ^{p + 1}}{p + 1} - \frac{( D + C a ) ^{p + 1}}{p + 1}}{C}

Vereinfache

= \frac{( D + C b ) ^{p + 1} - ( D + C a ) ^{p + 1}}{C ( p + 1 )}

\int_{e}^{e^{2}} \frac{( ln ( x ) ) ^{2}}{x} d x

\int_{4}^{b} x^{5} d x

\int_{1}^{e^{8}} \frac{8 ln ( x )}{x} d x

\int_{5}^{x} p^{2} sin (p) d p

\int_{- 2}^{- 1} 2 x^{2} - 4 x - 6 d x