integral from 3 to 4 of t^2-t-6

解答

\int_{3}^{4} t^{2} - t - 6 d t

\frac{17}{6}

十进制

2.83333 \dots

求解步骤

\int_{3}^{4} t^{2} - t - 6 d t

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}^{4} t^{2} d t - \int_{3}^{4} t d t - \int_{3}^{4} 6 d t

\int_{3}^{4} t^{2} d t = \frac{37}{3}

\int_{3}^{4} t d t = \frac{7}{2}

\int_{3}^{4} 6 d t = 6

= \frac{37}{3} - \frac{7}{2} - 6

化简

\frac{37}{3} - \frac{7}{2} - 6 : \frac{17}{6}

= \frac{17}{6}

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