integral from 0 to 5 of 8/(sqrt(7x+6))

Lösung

\int_{0}^{5} \frac{8}{7 x + 6} d x

\frac{8 ( 2 41 - 2 6 )}{7}

Dezimale

9.03687 \dots

Schritte zur Lösung

\int_{0}^{5} \frac{8}{7 x + 6} d x

Entferne die Konstante:

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

= 8 \cdot \int_{0}^{5} \frac{1}{7 x + 6} d x

Wende U-Substitution an

= 8 \cdot \int_{6}^{41} \frac{1}{7 u} d u

Entferne die Konstante:

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

= 8 \cdot \frac{1}{7} \cdot \int_{6}^{41} \frac{1}{u} d u

Wende die Potenzregel an

= 8 \cdot \frac{1}{7} [2 u^{\frac{1}{2}}]_{6}^{41}

Vereinfache

8 \cdot \frac{1}{7} [2 u^{\frac{1}{2}}]_{6}^{41} : \frac{8}{7} [2 u]_{6}^{41}

= \frac{8}{7} [2 u]_{6}^{41}

Berechne die Grenzen:

241 - 26

= \frac{8}{7} (241 - 26)

Vereinfache

= \frac{8 ( 2 41 - 2 6 )}{7}

\int_{0}^{10} - x^{2} + 10 x d x

\int_{1}^{4} 2^{2 x} d x

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

\int_{0}^{17 π} sin (x) d x

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