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

解

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

\frac{9}{2}

十進法表記

4.5

解答ステップ

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

簡素化

= \int_{1}^{4} - x^{2} + 5 x - 4 d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{1}^{4} x^{2} d x + \int_{1}^{4} 5 x d x - \int_{1}^{4} 4 d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 5 x d x = \frac{75}{2}

\int_{1}^{4} 4 d x = 12

= - 21 + \frac{75}{2} - 12

簡素化

- 21 + \frac{75}{2} - 12 : \frac{9}{2}

= \frac{9}{2}