integral from 5 to 9 of 1/(sqrt(9-x))

Lösung

\int_{5}^{9} \frac{1}{9 - x} d x

4

Schritte zur Lösung

\int_{5}^{9} \frac{1}{9 - x} d x

Wende U-Substitution an

= \int_{4}^{0} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{4} - \frac{1}{u} d u

Entferne die Konstante:

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

= - (- \int_{0}^{4} \frac{1}{u} d u)

Wende die Potenzregel an

= - (- [2 u^{\frac{1}{2}}]_{0}^{4})

Vereinfache

- (- [2 u^{\frac{1}{2}}]_{0}^{4}) : [2 u]_{0}^{4}

= [2 u]_{0}^{4}

Berechne die Grenzen:

4

= 4

\int_{0}^{2 π} cos^{5} (x) d x

\int_{0}^{1} \frac{1}{x ^{2} + 2 x + 5} d x

\int_{2}^{3} \frac{6}{2 x ^{2} + 2 x - 4} d x

\int_{0}^{1} 3 x^{2} x^{3} - 5 d x

\int_{1}^{5} \frac{x - 1}{x ^{2} ( x + 1 )} d x