integral from 0 to 25 of 1/(sqrt(25-x))

解答

\int_{0}^{25} \frac{1}{25 - x} d x

10

求解步骤

\int_{0}^{25} \frac{1}{25 - x} d x

使用换元积分法

= \int_{25}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{25} - \frac{1}{u} d u

提出常数:

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

= - (- \int_{0}^{25} \frac{1}{u} d u)

使用幂法则

= - (- [2 u^{\frac{1}{2}}]_{0}^{25})

化简

- (- [2 u^{\frac{1}{2}}]_{0}^{25}) : [2 u]_{0}^{25}

= [2 u]_{0}^{25}

带入上下限计算后得:

10

= 10

\int 12 x e^{6 x} d x

\frac{\partial}{\partial y} (x e^{y} + y)

\frac{d y}{d x} = x y^{\frac{1}{2}}

\frac{d}{d x} (2 ln (2))

\int \frac{( x ^{2} - 25 ) ^{\frac{3}{2}}}{x ^{6}} d x