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

Lösung

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

π

Dezimale

3.14159 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{2}} 16 sin^{2} (u) cos^{2} (u) d u

Entferne die Konstante:

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

= 16 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 16 \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 u )}{8} d u

Entferne die Konstante:

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

= 16 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 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

= 16 \cdot \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (4 u) d u)

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

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

= 16 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Vereinfache

16 \cdot \frac{1}{8} (\frac{π}{2} - 0) : π

= π

t an g e n t f (x) = \frac{- 8 x}{x ^{2} + 1}

l a pl a ce t r an s f or m 3 t + 2

x \to - 2 lim (x^{2} + 2 x - 3)

t an g e n t 4 cos (x)

2 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}