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人気のある微分積分 >

integral from-1 to 2 of x+|x|

解

\int_{- 1}^{2} x + ∣ x ∣ d x

解

4

解答ステップ

\int_{- 1}^{2} x + ∣ x ∣ d x

絶対数を排除する

= \int_{- 1}^{0} x - x d x + \int_{0}^{2} x + x d x

\int_{- 1}^{0} x - x d x = 0

\int_{0}^{2} x + x d x = 4

= 0 + 4

簡素化

= 4

グラフ

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