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

Lösung

\int_{- 2}^{2} (x^{2} 4 - x^{2}) d x

2 π

Dezimale

6.28318 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{- \frac{π}{2}}^{\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_{- \frac{π}{2}}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

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

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

= 16 \cdot \frac{1}{8} (π - 0)

Vereinfache

16 \cdot \frac{1}{8} (π - 0) : 2 π

= 2 π

\int_{- 3}^{2} x^{3} + x^{2} - 6 x d x

\int_{1}^{8} \frac{2}{x} d x

\int_{x}^{1} (1 - y) cos^{2} (z - y) d z

\int_{0}^{π} \frac{1}{2} (2 θ)^{2} d θ

\int_{1}^{9} \frac{3}{t} d t