integral de 0 a r de (br^2+cr)2pir

Solução

\int_{0}^{r} (b \cdot r^{2} + c \cdot r) 2 π r d r

\frac{π}{6} (3 b r^{4} + 4 c r^{3})

Passos da solução

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

Remover a constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Multiplicar pelo conjugado de

b r^{2} + cr : \frac{b ^{2} r ^{4} - c ^{2} r ^{2}}{b r ^{2} - cr}

= 2 π \cdot \int_{0}^{r} \frac{b ^{2} r ^{4} - c ^{2} r ^{2}}{b r ^{2} - cr} r d r

Simplificar

\frac{b ^{2} r ^{4} - c ^{2} r ^{2}}{b r ^{2} - cr} r : r^{2} (b r + c)

= 2 π \cdot \int_{0}^{r} r^{2} (b r + c) d r

Aplicar integração por partes

= 2 π [\frac{1}{3} r^{3} (b r + c) - \int \frac{b r ^{3}}{3} d r]_{0}^{r}

\int \frac{b r ^{3}}{3} d r = \frac{b r ^{4}}{12}

= 2 π [\frac{1}{3} r^{3} (b r + c) - \frac{b r ^{4}}{12}]_{0}^{r}

Simplificar

2 π [\frac{1}{3} r^{3} (b r + c) - \frac{b r ^{4}}{12}]_{0}^{r} : 2 π [\frac{1}{12} (3 b r^{4} + 4 c r^{3})]_{0}^{r}

= 2 π [\frac{1}{12} (3 b r^{4} + 4 c r^{3})]_{0}^{r}

Calcular os limites:

\frac{1}{12} (3 b r^{4} + 4 c r^{3})

= 2 π \frac{1}{12} (3 b r^{4} + 4 c r^{3})

Simplificar

= \frac{π}{6} (3 b r^{4} + 4 c r^{3})

\int_{0}^{3} x \cdot e^{x} d x

\int_{0}^{\frac{π}{4}} e^{x} cos (x) d x

\int_{0}^{3} \frac{5}{x ^{2} + 1} d x

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

\int_{- 1}^{2} (1 - ∣ x ∣) d x