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

Lösung

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

\frac{π}{8}

Dezimale

0.39269 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{1} 2 u^{2} 1 - u^{2} d u

Entferne die Konstante:

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

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

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 2 \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 v )}{8} d v

Entferne die Konstante:

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

= 2 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 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

= 2 \cdot \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d v - \int_{0}^{\frac{π}{2}} cos (4 v) d v)

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

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

= 2 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Vereinfache

2 \cdot \frac{1}{8} (\frac{π}{2} - 0) : \frac{π}{8}

= \frac{π}{8}

\int_{0}^{2} r e^{- r^{2}} d r

\int_{0}^{π} - 2 sin (x) d x

\int_{1}^{3} \frac{15 + 2 x ^{3}}{x ^{4}} d x

\int_{0}^{2} 1 - ∣ x - 1 ∣ d x

\int_{0}^{\frac{π}{2}} cos (x) \cdot sin (x) d x