integral from-1 to 5 of sqrt(5+4x-x^2)

Lösung

\int_{- 1}^{5} 5 + 4 x - x^{2} d x

\frac{9 π}{2}

Dezimale

14.13716 \dots

Schritte zur Lösung

\int_{- 1}^{5} 5 + 4 x - x^{2} d x

Vervollständige das Quadrat

5 + 4 x - x^{2} : - (x - 2)^{2} + 9

= \int_{- 1}^{5} - (x - 2)^{2} + 9 d x

Wende U-Substitution an

= \int_{- 3}^{3} - u^{2} + 9 d u

Trigonometrische Substitution anwenden

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 9 cos^{2} (v) d v

Entferne die Konstante:

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

= 9 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 9 \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

= 9 \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

= 9 \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

= 9 \cdot \frac{1}{2} (π + 0)

Vereinfache

9 \cdot \frac{1}{2} (π + 0) : \frac{9 π}{2}

= \frac{9 π}{2}

\int_{0}^{6} 4 d x

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

\int_{1}^{e} 2 π x ln (x) d x

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

\int_{0}^{0.5} 3 x^{2} d x