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

解答

\int_{1}^{4} (2 x^{2} - x + 1) d x

\frac{75}{2}

十进制

37.5

求解步骤

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

\int_{1}^{4} 2 x^{2} d x = 42

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} 1 d x = 3

= 42 - \frac{15}{2} + 3

化简

42 - \frac{15}{2} + 3 : \frac{75}{2}

= \frac{75}{2}

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