integral from 3.3 to 8.3 of 12x^2-6x+4/x

解答

\int_{3.3}^{8.3} 12 x^{2} - 6 x + \frac{4}{x} d x

1973.08933 \dots

求解步骤

\int_{3.3}^{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_{3.3}^{8.3} 12 x^{2} d x - \int_{3.3}^{8.3} 6 x d x + \int_{3.3}^{8.3} \frac{4}{x} d x

\int_{3.3}^{8.3} 12 x^{2} d x = 2143.39999 \dots

\int_{3.3}^{8.3} 6 x d x = 174

\int_{3.3}^{8.3} \frac{4}{x} d x = 3.68933 \dots

= 2143.39999 \dots - 174 + 3.68933 \dots

化简

= 1973.08933 \dots

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