integral from 1 to 2 of (x^2)/2+x/3-1/4

解答

\int_{1}^{2} \frac{x ^{2}}{2} + \frac{x}{3} - \frac{1}{4} d x

\frac{17}{12}

十进制

1.41666 \dots

求解步骤

\int_{1}^{2} \frac{x ^{2}}{2} + \frac{x}{3} - \frac{1}{4} 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_{1}^{2} \frac{x ^{2}}{2} d x + \int_{1}^{2} \frac{x}{3} d x - \int_{1}^{2} \frac{1}{4} d x

\int_{1}^{2} \frac{x ^{2}}{2} d x = \frac{7}{6}

\int_{1}^{2} \frac{x}{3} d x = \frac{1}{2}

\int_{1}^{2} \frac{1}{4} d x = \frac{1}{4}

= \frac{7}{6} + \frac{1}{2} - \frac{1}{4}

化简

\frac{7}{6} + \frac{1}{2} - \frac{1}{4} : \frac{17}{12}

= \frac{17}{12}

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