integral from 0 to 2 of sqrt(3x+1)

Lösung

\int_{0}^{2} 3 x + 1 d x

\frac{14 7 - 2}{9}

Dezimale

3.89339 \dots

Schritte zur Lösung

\int_{0}^{2} 3 x + 1 d x

Wende U-Substitution an

= \int_{1}^{7} \frac{u}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{1}^{7} u d u

Wende die Potenzregel an

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{7}

Berechne die Grenzen:

\frac{14 7}{3} - \frac{2}{3}

= \frac{1}{3} (\frac{14 7}{3} - \frac{2}{3})

Vereinfache

= \frac{14 7 - 2}{9}

\int_{0}^{3} (x^{2} + 5) d x

\int_{- π}^{0} - x sin (n x) d x

\int_{- π}^{0} (t - π) sin (n t) d t

\int_{2}^{9} \frac{1}{8 x + 9} d x

\int_{- 1}^{0} e^{- 2 x} d x