integral from 0 to 8 of sqrt(64-x^2)

Lösung

\int_{0}^{8} 64 - x^{2} d x

16 π

Dezimale

50.26548 \dots

Schritte zur Lösung

\int_{0}^{8} 64 - x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{2}} 64 cos^{2} (u) d u

Entferne die Konstante:

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

= 64 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 64 \cdot \int_{0}^{\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

= 64 \cdot \frac{1}{2} \cdot \int_{0}^{\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

= 64 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

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

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

= 64 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Vereinfache

64 \cdot \frac{1}{2} (\frac{π}{2} + 0) : 16 π

= 16 π

(x^{2} + 1) (\frac{d y}{d x}) + 8 x (y - 1) = 0, y (0) = 5

6 x + \frac{d y}{d x} = 0

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\int_{10}^{13} 2 x d x