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

Lösung

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

π

Dezimale

3.14159 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

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

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (2 x) d x = 0

= \frac{1}{2} (2 π + 0)

Vereinfache

\frac{1}{2} (2 π + 0) : π

= π

\int_{- \infty}^{- 3} \frac{x}{4 - x ^{2}} d x

\int_{0}^{4} 3 x^{3} d x

\int_{6.3}^{6.6} 12 x^{2} - 6 x + 4 d x

\int_{- 4}^{3} (- x^{2} - 3 x) d x

\int_{0}^{1} x^{4} ln (x) d x