integral from-1 to 1 of (x+1)

Lösung

\int_{- 1}^{1} (x + 1) d x

2

Schritte zur Lösung

\int_{- 1}^{1} x + 1 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_{- 1}^{1} x d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} x d x = 0

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

= 0 + 2

Vereinfache

= 2

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

\int_{- 1}^{0} 4 x^{2} 4 x^{3} + 6 d x

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

\int_{3}^{6} \frac{1}{x ^{2}} d x

\int_{π}^{2 π} sin (\frac{n x}{2}) d x