integral from 0 to pi of (1-cos(2x))

Lösung

\int_{0}^{π} (1 - cos (2 x)) d x

π

Dezimale

3.14159 \dots

Schritte zur Lösung

\int_{0}^{π} 1 - cos (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} 1 d x - \int_{0}^{π} cos (2 x) d x

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (2 x) d x = 0

= π - 0

Vereinfache

= π

\int_{0}^{2} \frac{1}{( 2 - x ) ^{2}} d x

\int_{0}^{\frac{5}{4}} 1 + 9 x d x

\int_{4}^{4.5} \frac{6 - x}{6} d x

\int_{51}^{100} \frac{1}{x} d x

\int_{0}^{π} sin^{2} (θ) sin (θ) d θ