integral from 1/2 to 3 of 4xln(2x)

解答

\int_{\frac{1}{2}}^{3} 4 x ln (2 x) d x

4 (- \frac{35}{16} + \frac{9}{2} ln (6))

十进制

23.50167 \dots

求解步骤

\int_{\frac{1}{2}}^{3} 4 x ln (2 x) d x

提出常数:

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

= 4 \cdot \int_{\frac{1}{2}}^{3} x ln (2 x) d x

使用分布积分法

= 4 [\frac{1}{2} x^{2} ln (2 x) - \int \frac{x}{2} d x]_{\frac{1}{2}}^{3}

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= 4 [\frac{1}{2} x^{2} ln (2 x) - \frac{x ^{2}}{4}]_{\frac{1}{2}}^{3}

带入上下限计算后得:

- \frac{35}{16} + \frac{9}{2} ln (6)

= 4 (- \frac{35}{16} + \frac{9}{2} ln (6))

\frac{\partial}{\partial x} (x^{2} \cdot sin (2 y))

x \to 2 lim (x^{2} - 6 x)

y^{^{''}} + 5 y^{^{'}} + 4 y = - 18 t e^{4 t}

\frac{d}{d x} (x e^{- s i n (x)} - x)

x \to 2 lim (2.1 - 1 + 3)