integral from 0 to a of 7x^2sqrt(a^2-x^2)

Lösung

\int_{0}^{a} 7 x^{2} a^{2} - x^{2} d x

\frac{7 π}{16} a^{4}

Schritte zur Lösung

\int_{0}^{a} 7 x^{2} a^{2} - x^{2} d x

Entferne die Konstante:

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

= 7 \cdot \int_{0}^{a} x^{2} a^{2} - x^{2} d x

Trigonometrische Substitution anwenden

= 7 \cdot \int_{0}^{\frac{π}{2}} a^{4} sin^{2} (u) cos^{2} (u) d u

Entferne die Konstante:

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

= 7 a^{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 7 a^{4} \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 u )}{8} d u

Entferne die Konstante:

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

= 7 a^{4} \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 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

= 7 a^{4} \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (4 u) d u)

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

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

= 7 a^{4} \frac{1}{8} (\frac{π}{2} - 0)

Vereinfache

7 a^{4} \frac{1}{8} (\frac{π}{2} - 0) : \frac{7 π}{16} a^{4}

= \frac{7 π}{16} a^{4}

\frac{d}{d x} (4 x^{2} - 8 x)

x \to 16 lim (\frac{x - 4}{4 x - 2})

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

d er i v a t i v e - 4 sin (x) - 9 ln (x) + 11

d er i v a t i v e f (x) = cos (\frac{x + 1}{x - 1})