integral from 0 to 6 of x/(sqrt(25+4x))

解答

\int_{0}^{6} \frac{x}{25 + 4 x} d x

\frac{17}{6}

十进制

2.83333 \dots

求解步骤

\int_{0}^{6} \frac{x}{25 + 4 x} d x

使用换元积分法

= \int_{5}^{7} \frac{u ^{2} - 25}{8} d u

提出常数:

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

= \frac{1}{8} \cdot \int_{5}^{7} u^{2} - 25 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

= \frac{1}{8} (\int_{5}^{7} u^{2} d u - \int_{5}^{7} 25 d u)

\int_{5}^{7} u^{2} d u = \frac{218}{3}

\int_{5}^{7} 25 d u = 50

= \frac{1}{8} (\frac{218}{3} - 50)

化简

\frac{1}{8} (\frac{218}{3} - 50) : \frac{17}{6}

= \frac{17}{6}

\int_{0}^{1} x^{2} cos (n x) d x

\int_{0}^{2} \frac{1}{2 x + 1} d x

\int_{- 1}^{4} - x^{2} + 4 x - 8 d x

\int_{0}^{1} 2 x^{3} e^{x^{2}} d x

\int_{0}^{3} \frac{2}{9} x^{2} d x