integral from 1 to x of y^{b-1}

Lösung

\int_{1}^{x} y^{b - 1} d y

\frac{x ^{b} - 1}{b}

Schritte zur Lösung

\int_{1}^{x} y^{b - 1} d y

Wende die Potenzregel an

= [\frac{y ^{b}}{b}]_{1}^{x}

Berechne die Grenzen:

\frac{x ^{b}}{b} - \frac{1}{b}

= \frac{x ^{b}}{b} - \frac{1}{b}

Vereinfache

= \frac{x ^{b} - 1}{b}

\int_{0}^{1} 2 arccos (x) d x

\int_{0}^{a} sin (c x) sin (c x) d x

\int_{1}^{e} \frac{4 ( 1 + ln ( x ) )}{x} d x

\int_{- 3}^{3} 9 d x

\int_{0}^{1} (2 - t^{2}) d t