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

解答

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

\frac{16}{15}

十进制

1.06666 \dots

求解步骤

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

使用分布积分法

= [\frac{1}{4} x^{4} (2 - x)^{2} - \int - \frac{1}{2} x^{4} (- x + 2) d x]_{0}^{2}

\int - \frac{1}{2} x^{4} (- x + 2) d x = \frac{x ^{6}}{12} - \frac{x ^{5}}{5}

= [\frac{1}{4} x^{4} (2 - x)^{2} - (\frac{x ^{6}}{12} - \frac{x ^{5}}{5})]_{0}^{2}

化简

[\frac{1}{4} x^{4} (2 - x)^{2} - (\frac{x ^{6}}{12} - \frac{x ^{5}}{5})]_{0}^{2} : [\frac{1}{30} x^{4} (5 x^{2} - 24 x + 30)]_{0}^{2}

= [\frac{1}{30} x^{4} (5 x^{2} - 24 x + 30)]_{0}^{2}

带入上下限计算后得:

\frac{16}{15}

= \frac{16}{15}

x \to 0 lim (\frac{e ^{x}}{ln ( x )})

d er i v a t i v e \frac{3 x ^{2} + x}{- x ^{2} + 3 x - 3}

t an g e n t 17 - x^{2}

d er i v a t i v e x e^{x} csc (x)

\int 6 y^{2} + \frac{1}{24 y ^{2}} d y