integral from 0 to pi/2 of 9sin^3(θ)

Lösung

\int_{0}^{\frac{π}{2}} 9 sin^{3} (θ) d θ

6

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} 9 sin^{3} (θ) d θ

Entferne die Konstante:

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

= 9 \cdot \int_{0}^{\frac{π}{2}} sin^{3} (θ) d θ

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 9 \cdot \int_{0}^{\frac{π}{2}} (1 - cos^{2} (θ)) sin (θ) d θ

Wende U-Substitution an

= 9 \cdot \int_{1}^{0} - 1 + u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 9 (- \int_{0}^{1} - 1 + u^{2} 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 (- (- \int_{0}^{1} 1 d u + \int_{0}^{1} u^{2} d u))

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} u^{2} d u = \frac{1}{3}

= 9 (- (- 1 + \frac{1}{3}))

Vereinfache

9 (- (- 1 + \frac{1}{3})) : 6

= 6

\int_{0}^{\infty} e^{- 3} d x

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

\int_{0}^{π} 1 - cos (θ) d θ

\int_{0}^{1} x^{3} e^{x} d x

\int_{1}^{\infty} t e^{- t} d t