integral from-7 to 0 of 1/(sqrt(9-x))

解答

\int_{- 7}^{0} \frac{1}{9 - x} d x

2

求解步骤

\int_{- 7}^{0} \frac{1}{9 - x} d x

使用换元积分法

= \int_{16}^{9} - \frac{1}{u} d u

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

= - \int_{9}^{16} - \frac{1}{u} d u

提出常数:

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

= - (- \int_{9}^{16} \frac{1}{u} d u)

使用幂法则

= - (- [2 u^{\frac{1}{2}}]_{9}^{16})

化简

- (- [2 u^{\frac{1}{2}}]_{9}^{16}) : [2 u]_{9}^{16}

= [2 u]_{9}^{16}

带入上下限计算后得:

2

= 2

\frac{d}{d x} (\frac{x ^{5}}{x ^{3}})

\frac{d}{d x} (5 x^{2} cos (x) + x)

\frac{\partial}{\partial y} (\frac{( y + x )}{( y - x )})

t an g e n t f (x) = x^{2} - 2

2 \frac{d y}{d x} = x y (x + 1)