integral from 0 to 3 of f(t)

Lösung

\int_{0}^{3} f (t) d t

\frac{9 f}{2}

Schritte zur Lösung

\int_{0}^{3} f t d t

Entferne die Konstante:

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

= f \cdot \int_{0}^{3} t d t

Wende die Potenzregel an

= f [\frac{t ^{2}}{2}]_{0}^{3}

Berechne die Grenzen:

\frac{9}{2}

= f \frac{9}{2}

Vereinfache

= \frac{9 f}{2}

\int_{2}^{4} 4 d x

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

\int_{1}^{3} 3^{4 x} d x

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

\int_{0}^{\frac{π}{2}} 2 x cos (x) d x