integral from 1 to 5 of sqrt(-x^2+6x-5)

Lösung

\int_{1}^{5} - x^{2} + 6 x - 5 d x

2 π

Dezimale

6.28318 \dots

Schritte zur Lösung

\int_{1}^{5} - x^{2} + 6 x - 5 d x

Vervollständige das Quadrat

- x^{2} + 6 x - 5 : - (x - 3)^{2} + 4

= \int_{1}^{5} - (x - 3)^{2} + 4 d x

Wende U-Substitution an

= \int_{- 2}^{2} - u^{2} + 4 d u

Trigonometrische Substitution anwenden

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 4 cos^{2} (v) d v

Entferne die Konstante:

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

= 4 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 4 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \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

= 4 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 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

= 4 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 v) d v)

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

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

= 4 \cdot \frac{1}{2} (π + 0)

Vereinfache

4 \cdot \frac{1}{2} (π + 0) : 2 π

= 2 π

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