integral from 0 to 1 of 1/(sqrt(x(1-x)))

Lösung

\int_{0}^{1} \frac{1}{x ( 1 - x )} d x

π

Dezimale

3.14159 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{1}{x ( 1 - x )} d x

Multipliziere aus

x (1 - x) : x - x^{2}

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

Vervollständige das Quadrat

x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{1}{4}

= \int_{0}^{1} \frac{1}{- ( x - \frac{1}{2} ) ^{2} + \frac{1}{4}} d x

Wende U-Substitution an

= \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{2}{- 4 u ^{2} + 1} d u

Entferne die Konstante:

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

= 2 \cdot \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{- 4 u ^{2} + 1} d u

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Integral einer Konstanten :

\int a d x = a x

= 2 \cdot \frac{1}{2} [1 \cdot v]_{- \frac{π}{2}}^{\frac{π}{2}}

Vereinfache

2 \cdot \frac{1}{2} [1 \cdot v]_{- \frac{π}{2}}^{\frac{π}{2}} : [v]_{- \frac{π}{2}}^{\frac{π}{2}}

= [v]_{- \frac{π}{2}}^{\frac{π}{2}}

Berechne die Grenzen:

π

= π