integral from 0 to 1 of sqrt(4-x^2)

Lösung

\int_{0}^{1} 4 - x^{2} d x

2 (\frac{π}{6} + \frac{3}{4})

Dezimale

1.91322 \dots

Schritte zur Lösung

\int_{0}^{1} 4 - x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{6}} 4 cos^{2} (u) d u

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{\frac{π}{6}} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 4 \cdot \int_{0}^{\frac{π}{6}} \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

= 4 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} 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

= 4 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{6}} 1 d u + \int_{0}^{\frac{π}{6}} cos (2 u) d u)

\int_{0}^{\frac{π}{6}} 1 d u = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 u) d u = \frac{3}{4}

= 4 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4})

Vereinfache

4 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4}) : 2 (\frac{π}{6} + \frac{3}{4})

= 2 (\frac{π}{6} + \frac{3}{4})

\int \frac{9}{x ^{4}} d x

a re a 7 x - x^{2}, 2 x

d er i v a t i v e f (x) = \frac{1}{2} (x + 1)^{- \frac{1}{2}}

d er i v a t i v e e^{c o s (3 x)}

y^{^{''}} + 4 y^{^{'}} - y = 0