integral from 0 to 1 of 2sqrt(t)

Lösung

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

\frac{4}{3}

Dezimale

1.33333 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende die Potenzregel an

= 2 [\frac{2}{3} t^{\frac{3}{2}}]_{0}^{1}

Berechne die Grenzen:

\frac{2}{3}

= 2 \cdot \frac{2}{3}

Vereinfache

= \frac{4}{3}

\int_{0}^{\frac{π}{3}} sin^{6} (x) d x

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

\int_{- 2}^{5} - \frac{5}{x ^{2}} d x

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

\int_{0}^{5} e^{- s t} (2 t) d t