integral from 0 to r of br^3+cr^2

Lösung

\int_{0}^{r} b \cdot r^{3} + c \cdot r^{2} d r

\frac{b r ^{4}}{4} + \frac{c r ^{3}}{3}

Schritte zur Lösung

\int_{0}^{r} b r^{3} + c r^{2} 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^{3} d r + \int_{0}^{r} c r^{2} d r

\int_{0}^{r} b r^{3} d r = \frac{b r ^{4}}{4}

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

= \frac{b r ^{4}}{4} + \frac{c r ^{3}}{3}

\int_{0}^{2} \frac{π}{2} r d r

\int_{1}^{3} 2 x - x^{2} d x

\int_{0}^{0.5} 4 x d x

\int_{0}^{π} (\frac{sin ( x )}{x}) d x

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