integral from 0 to 19/6 of pi(19/6-x)^2

Lösung

\int_{0}^{\frac{19}{6}} π (\frac{19}{6} - x)^{2} d x

\frac{6859 π}{648}

Dezimale

33.25337 \dots

Schritte zur Lösung

\int_{0}^{\frac{19}{6}} π (\frac{19}{6} - x)^{2} d x

Entferne die Konstante:

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

= π \cdot \int_{0}^{\frac{19}{6}} (\frac{19}{6} - x)^{2} d x

Wende U-Substitution an

= π \cdot \int_{\frac{19}{6}}^{0} - u^{2} d u

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

= π (- \int_{0}^{\frac{19}{6}} - u^{2} d u)

Entferne die Konstante:

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

= π (- (- \int_{0}^{\frac{19}{6}} u^{2} d u))

Wende die Potenzregel an

= π (- (- [\frac{u ^{3}}{3}]_{0}^{\frac{19}{6}}))

Vereinfache

= π [\frac{u ^{3}}{3}]_{0}^{\frac{19}{6}}

Berechne die Grenzen:

\frac{6859}{648}

= π \frac{6859}{648}

Vereinfache

= \frac{6859 π}{648}

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