integral from-1 to 4 of x/(sqrt(x+5))

解答

\int_{- 1}^{4} \frac{x}{x + 5} d x

\frac{8}{3}

十进制

2.66666 \dots

求解步骤

\int_{- 1}^{4} \frac{x}{x + 5} d x

使用换元积分法

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

提出常数:

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

= 2 \cdot \int_{2}^{3} u^{2} - 5 d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{2}^{3} u^{2} d u = \frac{19}{3}

\int_{2}^{3} 5 d u = 5

= 2 (\frac{19}{3} - 5)

化简

2 (\frac{19}{3} - 5) : \frac{8}{3}

= \frac{8}{3}

\int_{- 12}^{12} (144 - x^{2}) d x

\int_{0}^{1} (x + e^{x}) d x

\int_{1}^{5} \frac{( x - 1 ) ^{3}}{x ^{2}} d x

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

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