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人気のある微分積分 >

integral from 0 to 8 of (12)/(sqrt(x+1))

解

\int_{0}^{8} \frac{12}{x + 1} d x

解

48

解答ステップ

\int_{0}^{8} \frac{12}{x + 1} d x

定数を除く:

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

= 12 \cdot \int_{0}^{8} \frac{1}{x + 1} d x

u置換積分法を適用する

= 12 \cdot \int_{1}^{3} 2 d u

定数の積分:

\int a d x = a x

= 12 [2 u]_{1}^{3}

限度を計算する:

4

= 12 \cdot 4

簡素化

= 48

グラフ

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