integral from x to 1 of y^2-2y+1

解

\int_{x}^{1} y^{2} - 2 y + 1 d y

x^{2} - x + \frac{- x ^{3} + 1}{3}

解答ステップ

\int_{x}^{1} y^{2} - 2 y + 1 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_{x}^{1} y^{2} d y - \int_{x}^{1} 2 y d y + \int_{x}^{1} 1 d y

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

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

\int_{x}^{1} 1 d y = 1 - x

= \frac{1 - x ^{3}}{3} - (1 - x^{2}) + 1 - x

簡素化

= x^{2} - x + \frac{- x ^{3} + 1}{3}