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

Lösung

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

32 (arcsin (\frac{x}{8}) + \frac{1}{2} sin (2 arcsin (\frac{x}{8})))

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{0}^{a r c s i n (\frac{x}{8})} 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}^{a r c s i n (\frac{x}{8})} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 64 \cdot \int_{0}^{a r c s i n (\frac{x}{8})} \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}^{a r c s i n (\frac{x}{8})} 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}^{a r c s i n (\frac{x}{8})} 1 d u + \int_{0}^{a r c s i n (\frac{x}{8})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{x}{8})} 1 d u = arcsin (\frac{x}{8})

\int_{0}^{a r c s i n (\frac{x}{8})} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (\frac{x}{8}))

= 64 \cdot \frac{1}{2} (arcsin (\frac{x}{8}) + \frac{1}{2} sin (2 arcsin (\frac{x}{8})))

Vereinfache

64 \cdot \frac{1}{2} (arcsin (\frac{x}{8}) + \frac{1}{2} sin (2 arcsin (\frac{x}{8}))) : 32 (arcsin (\frac{x}{8}) + \frac{1}{2} sin (2 arcsin (\frac{x}{8})))

= 32 (arcsin (\frac{x}{8}) + \frac{1}{2} sin (2 arcsin (\frac{x}{8})))

\int_{2}^{5} ln (5 x) d x

\int_{0}^{1} 3 y d y

\int_{0}^{3} 4 x - x^{2} - x d x

\int_{0}^{2} (t - 2) d t

\int_{0}^{\infty} e^{- a x} x d x