integral from-1 to 3 of sqrt(9-x^2)

Lösung

\int_{- 1}^{3} 9 - x^{2} d x

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

Dezimale

10.01206 \dots

Schritte zur Lösung

\int_{- 1}^{3} 9 - x^{2} d x

Trigonometrische Substitution anwenden

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

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})}^{\frac{π}{2}} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

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})}^{\frac{π}{2}} 1 + cos (2 u) d u

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})}^{\frac{π}{2}} 1 d u + \int_{- a r c s i n (\frac{1}{3})}^{\frac{π}{2}} cos (2 u) d u)

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

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

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

Vereinfache

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

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

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