integral from 1 to 6 of (2x)/(sqrt(x+3))

Lösung

\int_{1}^{6} \frac{2 x}{x + 3} d x

\frac{40}{3}

Dezimale

13.33333 \dots

Schritte zur Lösung

\int_{1}^{6} \frac{2 x}{x + 3} d x

Entferne die Konstante:

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

= 2 \cdot \int_{1}^{6} \frac{x}{x + 3} d x

Wende U-Substitution an

= 2 \cdot \int_{2}^{3} 2 (u^{2} - 3) d u

Entferne die Konstante:

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

= 2 \cdot 2 \cdot \int_{2}^{3} u^{2} - 3 d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 2 (\int_{2}^{3} u^{2} d u - \int_{2}^{3} 3 d u)

\int_{2}^{3} u^{2} d u = \frac{19}{3}

\int_{2}^{3} 3 d u = 3

= 2 \cdot 2 (\frac{19}{3} - 3)

Vereinfache

2 \cdot 2 (\frac{19}{3} - 3) : \frac{40}{3}

= \frac{40}{3}