integral from 0 to 2-x of 6-3x-3y

解

\int_{0}^{2 - x} 6 - 3 x - 3 y d y

3 x^{2} - 12 x - \frac{3 ( - x + 2 ) ^{2}}{2} + 12

解答ステップ

\int_{0}^{2 - x} 6 - 3 x - 3 y 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_{0}^{2 - x} 6 d y - \int_{0}^{2 - x} 3 x d y - \int_{0}^{2 - x} 3 y d y

\int_{0}^{2 - x} 6 d y = 6 (- x + 2)

\int_{0}^{2 - x} 3 x d y = 3 x (- x + 2)

\int_{0}^{2 - x} 3 y d y = \frac{3 ( - x + 2 ) ^{2}}{2}

= 6 (- x + 2) - 3 x (- x + 2) - \frac{3 ( - x + 2 ) ^{2}}{2}

簡素化

6 (- x + 2) - 3 x (- x + 2) - \frac{3 ( - x + 2 ) ^{2}}{2} : 3 x^{2} - 12 x - \frac{3 ( - x + 2 ) ^{2}}{2} + 12

= 3 x^{2} - 12 x - \frac{3 ( - x + 2 ) ^{2}}{2} + 12