integral from-4 to 4 of (sqrt(16-x^2))

Lösung

\int_{- 4}^{4} (16 - x^{2}) d x

8 π

Dezimale

25.13274 \dots

Schritte zur Lösung

\int_{- 4}^{4} 16 - x^{2} d x

Trigonometrische Substitution anwenden

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 16 cos^{2} (u) d u

Entferne die Konstante:

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

= 16 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 16 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Entferne die Konstante:

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

= 16 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 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

= 16 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 u) d u)

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

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

= 16 \cdot \frac{1}{2} (π + 0)

Vereinfache

16 \cdot \frac{1}{2} (π + 0) : 8 π

= 8 π

y^{^{'}} = \frac{x}{x - 1}, y (2) = 5

\int \frac{x}{x ^{2} - 8 x} d x

\frac{d}{d x} (5 x sin (x))

d er i v a t i v e 3 x - 2

\frac{d}{d x} (3 x^{2} - 2 x + 4)