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

Lösung

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

\frac{π}{8}

Dezimale

0.39269 \dots

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Entferne die Konstante:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 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

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

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

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

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

Vereinfache

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

= \frac{π}{8}

\int (y - 1) (2 y + 1) d y

\frac{d y}{d x} + 2 x y = 2 x^{3}

in v erse l a pl a ce \frac{3 ( s + 3 )}{s ^{2} + 6 s + 8}

x \to \infty lim (3^{x} - x)

n = 1 \sum \infty \frac{n}{3 n + 2}