integral from 0 to 8 of x^2sqrt(64-x^2)

Lösung

\int_{0}^{8} x^{2} 64 - x^{2} d x

256 π

Dezimale

804.24771 \dots

Schritte zur Lösung

\int_{0}^{8} x^{2} 64 - x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{2}} 4096 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

= 4096 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 4096 \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

= 4096 \cdot \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

= 4096 \cdot \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

= 4096 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Vereinfache

4096 \cdot \frac{1}{8} (\frac{π}{2} - 0) : 256 π

= 256 π

y^{^{'}} - 5 y = 6 e^{7 t}

s l o p e 5 x - x^{2}

\int (- sin (x) + \frac{1}{cos ^{2} ( x )}) d x

a re a e^{2 x}, e^{x}, x = ln (3)

\frac{d}{d x} (x sin (\frac{9}{x}))