integral from 0 to 3 of 1/(sqrt(4-x))

Lösung

\int_{0}^{3} \frac{1}{4 - x} d x

2

Schritte zur Lösung

\int_{0}^{3} \frac{1}{4 - x} d x

Wende U-Substitution an

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

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

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

Entferne die Konstante:

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

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

Wende die Potenzregel an

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

Vereinfache

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

= [2 u]_{1}^{4}

Berechne die Grenzen:

2

= 2

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