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

解

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

\frac{4}{3}

十進法表記

1.33333 \dots

解答ステップ

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

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

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

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

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

簡素化

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

= \frac{4}{3}