integral from 0 to r of (br^2+cr)

Lösung

\int_{0}^{r} (b r^{2} + cr) d r

\frac{b r ^{3}}{3} + \frac{c r ^{2}}{2}

Schritte zur Lösung

\int_{0}^{r} b r^{2} + cr d r

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}^{r} b r^{2} d r + \int_{0}^{r} cr d r

\int_{0}^{r} b r^{2} d r = \frac{b r ^{3}}{3}

\int_{0}^{r} cr d r = \frac{c r ^{2}}{2}

= \frac{b r ^{3}}{3} + \frac{c r ^{2}}{2}

\int_{y - 2}^{2} (2) (- (y - x) + 2) d x

\int_{- 1}^{1} 1 3^{x} d x

\int_{- \infty}^{- 1} \frac{1}{x ^{5}} d x

\int_{2}^{4} \frac{1}{x ln ^{2} ( x )} d x

\int_{0}^{\infty} \frac{1}{y ^{3}} d y