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

Lösung

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

\frac{b sin ( c a )}{c} + d a

Schritte zur Lösung

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

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

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

= \frac{b sin ( c a )}{c} + d a

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

\int_{- 2}^{2} \frac{3 x y ^{2}}{x ^{2} + 1} d y

\int_{1}^{3} sin (r^{2}) r d r

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

\int_{x}^{1} y 1 + x^{2} - y^{2} d y