integral from 0 to 13 of sqrt(26x-x^2)

Lösung

\int_{0}^{13} 26 x - x^{2} d x

\frac{169 π}{4}

Dezimale

132.73228 \dots

Schritte zur Lösung

\int_{0}^{13} 26 x - x^{2} d x

Vervollständige das Quadrat

26 x - x^{2} : - (x - 13)^{2} + 169

= \int_{0}^{13} - (x - 13)^{2} + 169 d x

Wende U-Substitution an

= \int_{- 13}^{0} - u^{2} + 169 d u

Trigonometrische Substitution anwenden

= \int_{- \frac{π}{2}}^{0} 169 cos^{2} (v) d v

Entferne die Konstante:

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

= 169 \cdot \int_{- \frac{π}{2}}^{0} cos^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 169 \cdot \int_{- \frac{π}{2}}^{0} \frac{1 + cos ( 2 v )}{2} d v

Entferne die Konstante:

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

= 169 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{0} 1 + cos (2 v) 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

= 169 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{0} 1 d v + \int_{- \frac{π}{2}}^{0} cos (2 v) d v)

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

\int_{- \frac{π}{2}}^{0} cos (2 v) d v = 0

= 169 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Vereinfache

169 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{169 π}{4}

= \frac{169 π}{4}

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