integral from 1 to 6 of 1/((6-x)^{2/3)}

解答

\int_{1}^{6} \frac{1}{( 6 - x ) ^{\frac{2}{3}}} d x

3 \cdot 5^{\frac{1}{3}}

十进制

5.12992 \dots

求解步骤

\int_{1}^{6} \frac{1}{( 6 - x ) ^{\frac{2}{3}}} d x

使用换元积分法

= \int_{5}^{0} - \frac{1}{u ^{\frac{2}{3}}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{5} - \frac{1}{u ^{\frac{2}{3}}} d u

提出常数:

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

= - (- \int_{0}^{5} \frac{1}{u ^{\frac{2}{3}}} d u)

使用幂法则

= - (- [3 u^{\frac{1}{3}}]_{0}^{5})

化简

= [3 u^{\frac{1}{3}}]_{0}^{5}

带入上下限计算后得:

3 \cdot 5^{\frac{1}{3}}

= 3 \cdot 5^{\frac{1}{3}}

\int \frac{3}{2} (t + 1)^{\frac{1}{2}} d t

d er i v a t i v e \frac{x ^{3} + 7}{x}

\int x^{2} \cdot ln (2 x - 3) d x

x \to π lim (\frac{1 - cos ( x )}{x})

\frac{d}{d x} (\frac{2 x}{x ^{2} + 81})