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

Lösung

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

- π

Dezimale

- 3.14159 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

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

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

= - 4 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Vereinfache

- 4 \cdot \frac{1}{2} (\frac{π}{2} + 0) : - π

= - π

\int_{0}^{0.2} sin^{2} (x) d x

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

\int_{0}^{5} x^{3} - 1 d x

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

\int_{- 1}^{1} (- x^{2} - 2 x + 8) d x