integral from 2 to 1 of y^2-9

解

\int_{2}^{1} y^{2} - 9 d y

\frac{20}{3}

十進法表記

6.66666 \dots

解答ステップ

\int_{2}^{1} y^{2} - 9 d y

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{2} y^{2} - 9 d y

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}^{2} y^{2} d y - \int_{1}^{2} 9 d y)

\int_{1}^{2} y^{2} d y = \frac{7}{3}

\int_{1}^{2} 9 d y = 9

= - (\frac{7}{3} - 9)

簡素化

- (\frac{7}{3} - 9) : \frac{20}{3}

= \frac{20}{3}