integral from 0 to 3 of 1/9 x^2

Lösung

\int_{0}^{3} \frac{1}{9} x^{2} d x

1

Schritte zur Lösung

\int_{0}^{3} \frac{1}{9} x^{2} d x

Entferne die Konstante:

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

= \frac{1}{9} \cdot \int_{0}^{3} x^{2} d x

Wende die Potenzregel an

= \frac{1}{9} [\frac{x ^{3}}{3}]_{0}^{3}

Berechne die Grenzen:

9

= \frac{1}{9} \cdot 9

Vereinfache

= 1

\int_{0}^{π} (sin (x))^{4} d x

\int_{1}^{6} 1

\int_{0}^{9} \frac{4}{( 3 + x ) ^{2}} d x

\int_{0}^{1} x^{12} + 1 2^{x} d x

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