integral from-0.5 to 0 of b+2bx

Lösung

\int_{- 0.5}^{0} b + 2 b x d x

0.25 b

Schritte zur Lösung

\int_{- 0.5}^{0} b + 2 b 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_{- 0.5}^{0} b d x + \int_{- 0.5}^{0} 2 b x d x

\int_{- 0.5}^{0} b d x = 0.5 b

\int_{- 0.5}^{0} 2 b x d x = - 0.25 b

= 0.5 b - 0.25 b

Vereinfache

= 0.25 b

\int_{- 1}^{1} (x^{5} - 4 x^{2} + x - 1) d x

\int_{0}^{1} t^{2} + t d t

\int_{1}^{2} t \frac{1}{2} t^{2} + 6 d t

\int_{0}^{7} 2 x d x

\int_{- π}^{π} cos^{2} (k x) d x