integral from 0 to 2 of f(2x)

Lösung

\int_{0}^{2} f (2 x) d x

4 f

Schritte zur Lösung

\int_{0}^{2} f \cdot 2 x d x

Entferne die Konstante:

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

= 2 f \cdot \int_{0}^{2} x d x

Wende die Potenzregel an

= 2 f [\frac{x ^{2}}{2}]_{0}^{2}

Berechne die Grenzen:

2

= 2 f \cdot 2

Vereinfache

= 4 f

\int_{0}^{6} (x^{2} + 1) d x

\int_{0}^{2} \frac{1}{x + 4} d x

\int_{1}^{6} t^{2} - 2 t - 8 d t

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

\int_{x^{2}}^{x} (x^{2} + y^{2}) d y