integral from 0 to 1 of 4sqrt(1-x^2)

Lösung

\int_{0}^{1} 4 1 - x^{2} d x

π

Dezimale

3.14159 \dots

Schritte zur Lösung

\int_{0}^{1} 4 1 - x^{2} d x

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{1} 1 - x^{2} d x

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 4 \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

= 4 \cdot \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

= 4 \cdot \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

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

Vereinfache

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

= π

\int_{0}^{3} \frac{1}{x ^{2} - 8 x + 7} d x

\int_{- 3}^{3} (2 x + 6) d x

\int_{- 1}^{0} \frac{1}{x ^{2} - 2 x} d x

\int_{- 1}^{1} \frac{1}{x ^{2}} e^{\frac{2}{x}} d x

\int_{0}^{2 π} s^{2} d s