integral from 1 to 2 of (2x-1)^5

Lösung

\int_{1}^{2} (2 x - 1)^{5} d x

\frac{182}{3}

Dezimale

60.66666 \dots

Schritte zur Lösung

\int_{1}^{2} (2 x - 1)^{5} d x

Wende U-Substitution an

= \int_{1}^{3} \frac{u ^{5}}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{3} u^{5} d u

Wende die Potenzregel an

= \frac{1}{2} [\frac{u ^{6}}{6}]_{1}^{3}

Berechne die Grenzen:

\frac{364}{3}

= \frac{1}{2} \cdot \frac{364}{3}

Vereinfache

= \frac{182}{3}

\int_{0}^{6} (- x^{2} + 6 x) d x

\int_{0}^{1} \frac{e ^{x}}{x ^{2}} d x

\int_{2}^{x^{3}} \frac{1}{p ^{2}} d p

\int_{5}^{10} (x^{2} + 1)^{5} d x

\int_{0}^{2} t 4 t^{2} + 1 d t