integral from 2 to 4 of (sqrt(x-2))/x

Lösung

\int_{2}^{4} \frac{x - 2}{x} d x

22 (- \frac{π}{4} + 1)

Dezimale

0.60698 \dots

Schritte zur Lösung

\int_{2}^{4} \frac{x - 2}{x} d x

Wende U-Substitution an

= \int_{0}^{2} \frac{2 u ^{2}}{u ^{2} + 2} d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{2} \frac{u ^{2}}{u ^{2} + 2} d u

Wende integrale Substitution an

= 2 \cdot \int_{0}^{1} \frac{2 v ^{2}}{v ^{2} + 1} d v

Entferne die Konstante:

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

= 22 \cdot \int_{0}^{1} \frac{v ^{2}}{v ^{2} + 1} d v

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

= 22 \cdot \int_{0}^{1} - \frac{1}{v ^{2} + 1} + 1 d v

Wende die Summenregel an:

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

= 22 (- \int_{0}^{1} \frac{1}{v ^{2} + 1} d v + \int_{0}^{1} 1 d v)

\int_{0}^{1} \frac{1}{v ^{2} + 1} d v = \frac{π}{4}

\int_{0}^{1} 1 d v = 1

= 22 (- \frac{π}{4} + 1)

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