integral from 1 to e^3 of x^2ln(x)

解答

\int_{1}^{e^{3}} x^{2} ln (x) d x

\frac{8 e ^{9} + 1}{9}

十进制

7202.85238 \dots

求解步骤

\int_{1}^{e^{3}} x^{2} ln (x) d x

使用分布积分法

= [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \int \frac{x ^{2}}{3} d x)]_{1}^{e^{3}}

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

= [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \frac{x ^{3}}{9})]_{1}^{e^{3}}

3 \cdot \frac{x ^{3}}{9} = \frac{x ^{3}}{3}

= [\frac{1}{3} (x^{3} ln (x) - \frac{x ^{3}}{3})]_{1}^{e^{3}}

带入上下限计算后得:

\frac{8 e ^{9}}{9} + \frac{1}{9}

= \frac{8 e ^{9}}{9} + \frac{1}{9}

化简

= \frac{8 e ^{9} + 1}{9}

x \to 5 lim (123)

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

(- 1 x y^{2} + 2 y) d x + (- 1 x^{2} y + 2 x) d y = 0

d er i v a t i v e 4 sec (θ)

\frac{\partial}{\partial x} (x^{\frac{3}{2}} y)