integral from 1 to 3 of sqrt(5+4x-x^2)

Lösung

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

9 arcsin (\frac{1}{3}) + 22

Dezimale

5.88695 \dots

Schritte zur Lösung

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

Vervollständige das Quadrat

5 + 4 x - x^{2} : - (x - 2)^{2} + 9

= \int_{1}^{3} - (x - 2)^{2} + 9 d x

Wende U-Substitution an

= \int_{- 1}^{1} - u^{2} + 9 d u

Trigonometrische Substitution anwenden

= \int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} 9 cos^{2} (v) d v

Entferne die Konstante:

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

= 9 \cdot \int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} cos^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 9 \cdot \int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} \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

= 9 \cdot \frac{1}{2} \cdot \int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} 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

= 9 \cdot \frac{1}{2} (\int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} 1 d v + \int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} cos (2 v) d v)

\int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} 1 d v = 2 arcsin (\frac{1}{3})

\int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} cos (2 v) d v = \frac{4 2}{9}

= 9 \cdot \frac{1}{2} (2 arcsin (\frac{1}{3}) + \frac{4 2}{9})

Vereinfache

9 \cdot \frac{1}{2} (2 arcsin (\frac{1}{3}) + \frac{4 2}{9}) : 9 arcsin (\frac{1}{3}) + 22

= 9 arcsin (\frac{1}{3}) + 22

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