integral from 0 to z of sqrt(z^2-y^2)

Lösung

\int_{0}^{z} z^{2} - y^{2} d y

\frac{π}{4} z^{2}

Schritte zur Lösung

\int_{0}^{z} z^{2} - y^{2} d y

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= z^{2} \cdot \int_{0}^{\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

= z^{2} \frac{1}{2} \cdot \int_{0}^{\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

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

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

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

= z^{2} \frac{1}{2} (\frac{π}{2} + 0)

Vereinfache

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

= \frac{π}{4} z^{2}

\int_{1}^{2} x e^{6 x^{2}} d x

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