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人気のある微分積分 >

integral from 6 to 7 of 1/(sqrt(7-x))

解

\int_{6}^{7} \frac{1}{7 - x} d x

解

2

解答ステップ

\int_{6}^{7} \frac{1}{7 - x} d x

u置換積分法を適用する

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

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

= - \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

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

べき乗則を適用する

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

簡素化

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

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

限度を計算する:

2

= 2

グラフ

人気の例

integral from 1 to 3 of x^2*f(x^3)

\int_{1}^{3} x^{2} \cdot f (x^{3}) d x

integral from 0 to 2 of (-x+2)

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

integral from 0 to 2 of (x^2)/(x^3+7)

\int_{0}^{2} \frac{x ^{2}}{x ^{3} + 7} d x

integral from-1 to 0 of xln(1-x)

\int_{- 1}^{0} x ln (1 - x) d x

integral from 9 to 1 of 1/x

\int_{9}^{1} \frac{1}{x} d x