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

Lösung

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

π

Dezimale

3.14159 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

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

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} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

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 v) d v

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 v + \int_{- \frac{π}{2}}^{0} cos (2 v) d v)

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

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

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

Vereinfache

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

= π

\int_{0}^{5} 7 \cdot 3 x^{2} d x

\int_{0}^{2} (9 - t) t d t

\int_{4}^{5} (2 x - 7)^{4} d x

\int_{0}^{1} (7 x^{e} + 5 e^{x}) d x

\int_{- 3}^{1} g (x) d x