integral from 1 to 9 of xsqrt(2x+7)

解答

\int_{1}^{9} x 2 x + 7 d x

\frac{2608}{15}

十进制

173.86666 \dots

求解步骤

\int_{1}^{9} x 2 x + 7 d x

使用换元积分法

= \int_{9}^{25} \frac{u ( u - 7 )}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int_{9}^{25} u (u - 7) d u

乘开

u (u - 7) : u^{\frac{3}{2}} - 7 u

= \frac{1}{4} \cdot \int_{9}^{25} u^{\frac{3}{2}} - 7 u 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}{4} (\int_{9}^{25} u^{\frac{3}{2}} d u - \int_{9}^{25} 7 u d u)

\int_{9}^{25} u^{\frac{3}{2}} d u = \frac{5764}{5}

\int_{9}^{25} 7 u d u = \frac{1372}{3}

= \frac{1}{4} (\frac{5764}{5} - \frac{1372}{3})

化简

\frac{1}{4} (\frac{5764}{5} - \frac{1372}{3}) : \frac{2608}{15}

= \frac{2608}{15}

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