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人気のある微分積分 >

integral from 0 to 1 of 2(x+y)

解

\int_{0}^{1} 2 (x + y) d x

解

2 y + 1

解答ステップ

\int_{0}^{1} 2 (x + y) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

= 2 (\int_{0}^{1} y d x + \int_{0}^{1} x d x)

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

\int_{0}^{1} x d x = \frac{1}{2}

= 2 (y + \frac{1}{2})

簡素化

2 (y + \frac{1}{2}) : 2 y + 1

= 2 y + 1

人気の例

integral from 2 to 4 of x/((x+1)(x+2))

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