integral from 0 to pi of x+sin(x)

Lösung

\int_{0}^{π} x + sin (x) d x

\frac{π ^{2}}{2} + 2

Dezimale

6.93480 \dots

Schritte zur Lösung

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

\int_{0}^{π} x d x = \frac{π ^{2}}{2}

\int_{0}^{π} sin (x) d x = 2

= \frac{π ^{2}}{2} + 2

\int_{0}^{1} x^{2} e^{- 2 x} d x

\frac{d}{d x} (\frac{1}{x ^{2} + a ^{2}})

\int \frac{x ^{2} + 2 x + 1}{x ^{2}} d x

t a y l or sin (x), 3

x \to 2 lim (4 x + 1)