integral from 2.7 to 8.3 of 12x^2-6x+4/x

解答

\int_{2.7}^{8.3} 12 x^{2} - 6 x + \frac{4}{x} d x

2028.10801 \dots

求解步骤

\int_{2.7}^{8.3} 12 x^{2} - 6 x + \frac{4}{x} d x

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

= \int_{2.7}^{8.3} 12 x^{2} d x - \int_{2.7}^{8.3} 6 x d x + \int_{2.7}^{8.3} \frac{4}{x} d x

\int_{2.7}^{8.3} 12 x^{2} d x = 2208.41599 \dots

\int_{2.7}^{8.3} 6 x d x = 184.8

\int_{2.7}^{8.3} \frac{4}{x} d x = 4.49201 \dots

= 2208.41599 \dots - 184.8 + 4.49201 \dots

化简

= 2028.10801 \dots

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