integral from 0 to pi of 2sin(x)-sin(2x)

Lösung

\int_{0}^{π} 2 sin (x) - sin (2 x) d x

4

Schritte zur Lösung

\int_{0}^{π} 2 sin (x) - sin (2 x) d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} 2 sin (x) d x - \int_{0}^{π} sin (2 x) d x

\int_{0}^{π} 2 sin (x) d x = 4

\int_{0}^{π} sin (2 x) d x = 0

= 4 - 0

Vereinfache

= 4

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

\int_{π}^{2 π} cos (θ) d θ

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

\int_{- 1}^{1} (z + 1)^{5} d z

\int_{1}^{2} \frac{1}{x ^{2} - 6 x + 9} d x