integral from 0 to 2 of pi(1/(x+4))^2

Lösung

\int_{0}^{2} π (\frac{1}{x + 4})^{2} d x

\frac{π}{12}

Dezimale

0.26179 \dots

Schritte zur Lösung

\int_{0}^{2} π (\frac{1}{x + 4})^{2} d x

Entferne die Konstante:

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

= π \cdot \int_{0}^{2} (\frac{1}{x + 4})^{2} d x

Vereinfache

(\frac{1}{x + 4})^{2} : \frac{1}{( x + 4 ) ^{2}}

= π \cdot \int_{0}^{2} \frac{1}{( x + 4 ) ^{2}} d x

Wende U-Substitution an

= π \cdot \int_{4}^{6} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= π [- \frac{1}{u}]_{4}^{6}

Berechne die Grenzen:

\frac{1}{12}

= π \frac{1}{12}

Vereinfache

= \frac{π}{12}

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