integral from 0 to 1 of sqrt(1-z^2)

Lösung

\int_{0}^{1} 1 - z^{2} d z

\frac{π}{4}

Dezimale

0.78539 \dots

Schritte zur Lösung

\int_{0}^{1} 1 - z^{2} d z

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \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

= \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

= \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

= \frac{1}{2} (\frac{π}{2} + 0)

Vereinfache

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

= \frac{π}{4}

\int_{0}^{4} (3 - x) d x

\int_{- 2}^{4} x ∣ x - 3 ∣ d x

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

\int_{1}^{2} (z + \frac{2}{z}) \cdot (z - \frac{2}{z}) d z

\int_{0}^{1} x 35 x^{2} + 1 d x