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受欢迎的微积分 >

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

解答

\int_{0}^{r} (b r^{2} + cr) 2 π r d r

解答

\frac{π}{6} (3 b r^{4} + 4 c r^{3})

求解步骤

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

提出常数:

\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

乘以

b r^{2} + cr

的共轭根式

: \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

化简

\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

使用分布积分法

= 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}

化简

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}

带入上下限计算后得:

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

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

化简

= \frac{π}{6} (3 b r^{4} + 4 c r^{3})

流行的例子

integral from 0 to pi/3 of (cos(x)+sec(x))^2

\int_{0}^{\frac{π}{3}} (cos (x) + sec (x))^{2} d x

integral from 0 to 4 of sqrt(2-\sqrt{x)}

\int_{0}^{4} 2 - x d x

integral from 0 to 1 of x(x^2-1)^3

\int_{0}^{1} x (x^{2} - 1)^{3} d x

integral from 0 to 1 of x/(sqrt(5x^2+4))

\int_{0}^{1} \frac{x}{5 x ^{2} + 4} d x

integral from 1 to 5 of 3^{2x}

\int_{1}^{5} 3^{2 x} d x