integral from 1 to 2 of (-y^3+2y^2+5y-6)

解

\int_{1}^{2} (- y^{3} + 2 y^{2} + 5 y - 6) d y

\frac{29}{12}

十進法表記

2.41666 \dots

解答ステップ

\int_{1}^{2} - y^{3} + 2 y^{2} + 5 y - 6 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^{3} d y + \int_{1}^{2} 2 y^{2} d y + \int_{1}^{2} 5 y d y - \int_{1}^{2} 6 d y

\int_{1}^{2} y^{3} d y = \frac{15}{4}

\int_{1}^{2} 2 y^{2} d y = \frac{14}{3}

\int_{1}^{2} 5 y d y = \frac{15}{2}

\int_{1}^{2} 6 d y = 6

= - \frac{15}{4} + \frac{14}{3} + \frac{15}{2} - 6

簡素化

- \frac{15}{4} + \frac{14}{3} + \frac{15}{2} - 6 : \frac{29}{12}

= \frac{29}{12}