integral from 0 to 1 of 2z

Lösung

\int_{0}^{1} 2 z d z

1

Schritte zur Lösung

\int_{0}^{1} 2 z d z

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} z d z

Wende die Potenzregel an

= 2 [\frac{z ^{2}}{2}]_{0}^{1}

Berechne die Grenzen:

\frac{1}{2}

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

Vereinfache

= 1

\int_{1}^{2} (4 x^{3} - \frac{x ^{4}}{5}) d x

\int_{0}^{1} - 6 x 5^{x} d x

\int_{1}^{3} 3 x x - 1 d x

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

\int_{1}^{3} (2 x + 1) (3 - x) d x