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

Lösung

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

- 2 π

Dezimale

- 6.28318 \dots

Schritte zur Lösung

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= - 4 \cdot \int_{- \frac{π}{2}}^{\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

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

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

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

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

= - 4 \cdot \frac{1}{2} (π + 0)

Vereinfache

- 4 \cdot \frac{1}{2} (π + 0) : - 2 π

= - 2 π

\int_{0}^{1} sin^{4} (x) cos (x) d x

\int_{0}^{1} (2 x + x) d x

\int_{π}^{\frac{3 π}{2}} - sin (x) d x

\int_{- 1}^{2} 8 x - x^{2} - 7 d x

\int_{- 1}^{0} - 2 x + 3 d x