integral from 0 to 1 of (r^3)/(4+r^2)

解答

\int_{0}^{1} \frac{r ^{3}}{4 + r ^{2}} d r

\frac{1}{2} (1 - 4 (ln (5) - 2 ln (2)))

十进制

0.05371 \dots

求解步骤

\int_{0}^{1} \frac{r ^{3}}{4 + r ^{2}} d r

使用换元积分法

= \int_{4}^{5} \frac{u - 4}{2 u} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{4}^{5} \frac{u - 4}{u} d u

乘开

\frac{u - 4}{u} : 1 - \frac{4}{u}

= \frac{1}{2} \cdot \int_{4}^{5} 1 - \frac{4}{u} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{4}^{5} 1 d u - \int_{4}^{5} \frac{4}{u} d u)

\int_{4}^{5} 1 d u = 1

\int_{4}^{5} \frac{4}{u} d u = 4 (ln (5) - 2 ln (2))

= \frac{1}{2} (1 - 4 (ln (5) - 2 ln (2)))

x \to 2 lim (\frac{5 - x}{x - 2})

\frac{d}{d x} (\frac{sin ( x ) - x}{x ^{3}})

x \to \infty lim (4 e^{x})

\int_{π}^{\infty} sin (x) d x

\frac{d}{d x} (16 e^{- 4 x})