integral from-8 to 4 of sqrt(32-4x-x^2)

Lösung

\int_{- 8}^{4} 32 - 4 x - x^{2} d x

18 π

Dezimale

56.54866 \dots

Schritte zur Lösung

\int_{- 8}^{4} 32 - 4 x - x^{2} d x

Vervollständige das Quadrat

32 - 4 x - x^{2} : - (x + 2)^{2} + 36

= \int_{- 8}^{4} - (x + 2)^{2} + 36 d x

Wende U-Substitution an

= \int_{- 6}^{6} - u^{2} + 36 d u

Trigonometrische Substitution anwenden

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 36 cos^{2} (v) d v

Entferne die Konstante:

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

= 36 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 36 \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

= 36 \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

= 36 \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

= 36 \cdot \frac{1}{2} (π + 0)

Vereinfache

36 \cdot \frac{1}{2} (π + 0) : 18 π

= 18 π

\int_{- 1}^{1} x (x^{3} + cos^{3} (x)) d x

\int_{0}^{2 π} - 27 sin^{2} (t) d t

\int_{1}^{2} \frac{1}{( 3 - x ) ^{2}} d x

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

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