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

解

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

\frac{4}{3}

十進法表記

1.33333 \dots

解答ステップ

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

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 4 x d x = 16

\int_{1}^{3} 3 d x = 6

= - \frac{26}{3} + 16 - 6

簡素化

- \frac{26}{3} + 16 - 6 : \frac{4}{3}

= \frac{4}{3}