integral from 0 to 3 of sqrt(9-t^2)

Lösung

\int_{0}^{3} 9 - t^{2} d t

\frac{9 π}{4}

Dezimale

7.06858 \dots

Schritte zur Lösung

\int_{0}^{3} 9 - t^{2} d t

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{2}} 9 cos^{2} (u) d u

Entferne die Konstante:

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

= 9 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 9 \cdot \int_{0}^{\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

= 9 \cdot \frac{1}{2} \cdot \int_{0}^{\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

= 9 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

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

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

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

Vereinfache

9 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{9 π}{4}

= \frac{9 π}{4}

\int_{0}^{\frac{1}{4}} 2 x d x

\int_{0}^{4} 3 x^{2} - x + 1 d x

\int_{0}^{2 π} z cos (n z) d z

\int_{0}^{1} 10 x^{2} d x

\int_{0}^{c o s (θ)} 3 r^{2} sin (θ) d r