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人気のある微分積分 >

integral from 0 to 3 of 1/(sqrt(4-x))

解

\int_{0}^{3} \frac{1}{4 - x} d x

解

2

解答ステップ

\int_{0}^{3} \frac{1}{4 - x} d x

u置換積分法を適用する

= \int_{4}^{1} - \frac{1}{u} d u

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

= - \int_{1}^{4} - \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_{1}^{4} \frac{1}{u} d u)

べき乗則を適用する

= - (- [2 u^{\frac{1}{2}}]_{1}^{4})

簡素化

- (- [2 u^{\frac{1}{2}}]_{1}^{4}) : [2 u]_{1}^{4}

= [2 u]_{1}^{4}

限度を計算する:

2

= 2

グラフ

人気の例

integral from-6 to 5 of |x|

\int_{- 6}^{5} ∣ x ∣ d x

integral from 0 to 1 of-2x^2+2x

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

integral from 0 to 2pi of (cos(x))^3

\int_{0}^{2 π} (cos (x))^{3} d x

integral from 1 to 25 of 1/(sqrt(x))

\int_{1}^{25} \frac{1}{x} d x

integral from 5 to 9 of 1/(x^2)

\int_{5}^{9} \frac{1}{x ^{2}} d x