integral from 0 to x of t^k

Lösung

\int_{0}^{x} t^{k} d t

\frac{x ^{k + 1}}{k + 1}

Schritte zur Lösung

\int_{0}^{x} t^{k} d t

Wende die Potenzregel an

= [\frac{t ^{k + 1}}{k + 1}]_{0}^{x}

Berechne die Grenzen:

\frac{x ^{k + 1}}{k + 1}

= \frac{x ^{k + 1}}{k + 1}

\int_{0}^{7} 1 + x^{3} d x

\int_{0}^{3} 3 x^{2} - 24 x + 45 d x

\int_{x^{2}}^{x} \frac{sin ( x t )}{t} d t

\int_{1}^{4} \frac{x - 2}{x ( x + 1 )} d x

\int_{a}^{b} \frac{1}{1 - x} d x