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人気のある微分積分 >

integral from 2 to 3 of (40)/(sqrt(3-x))

解

\int_{2}^{3} \frac{40}{3 - x} d x

解

80

解答ステップ

\int_{2}^{3} \frac{40}{3 - x} d x

定数を除く:

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

= 40 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

u置換積分法を適用する

= 40 \cdot \int_{1}^{0} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 40 (- \int_{0}^{1} - \frac{1}{u} d u)

定数を除く:

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

= 40 (- (- \int_{0}^{1} \frac{1}{u} d u))

べき乗則を適用する

= 40 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

簡素化

40 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 40 [2 u]_{0}^{1}

= 40 [2 u]_{0}^{1}

限度を計算する:

2

= 40 \cdot 2

簡素化

= 80

グラフ

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