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

Lösung

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

\frac{π}{2} + 1

Dezimale

2.57079 \dots

Schritte zur Lösung

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

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

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

= \frac{π}{2} + 1

\int_{0}^{2 π} sin (x) cos (n x) d x

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