integral from 0 to a of (bsin(cx)+

Lösung

\int_{0}^{a} (b sin (c x) + d) d x

\frac{b ( - cos ( c a ) + 1 )}{c} + d a

Schritte zur Lösung

\int_{0}^{a} b sin (c x) + dd 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}^{a} b sin (c x) d x + \int_{0}^{a} dd x

\int_{0}^{a} b sin (c x) d x = \frac{b ( - cos ( c a ) + 1 )}{c}

\int_{0}^{a} dd x = d a

= \frac{b ( - cos ( c a ) + 1 )}{c} + d a

\int_{0}^{1} \frac{1 + x ^{2}}{1 + x ^{4}} d x

\int_{100}^{s} \frac{1}{0.2 s} d s

\int_{0}^{2 π} 7 x 2 d x

\int_{0}^{1} e^{- 10 x} d x

\int_{- 1}^{0} x 1 + 9 x^{4} d x