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人気のある微分積分 >

integral from-2 to 0 of 1/(sqrt(1-4x))

解

\int_{- 2}^{0} \frac{1}{1 - 4 x} d x

解

1

解答ステップ

\int_{- 2}^{0} \frac{1}{1 - 4 x} d x

u置換積分法を適用する

= \int_{9}^{1} - \frac{1}{4 u} d u

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

= - \int_{1}^{9} - \frac{1}{4 u} d u

定数を除く:

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

= - (- \frac{1}{4} \cdot \int_{1}^{9} \frac{1}{u} d u)

べき乗則を適用する

= - (- \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{9})

簡素化

- (- \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{9}) : \frac{1}{4} [2 u]_{1}^{9}

= \frac{1}{4} [2 u]_{1}^{9}

限度を計算する:

4

= \frac{1}{4} \cdot 4

簡素化

= 1

グラフ

人気の例

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

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

integral from 0 to 1 of (x^2)/(x^3+2)

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 2} d x

integral from 0 to 1 of 2xsqrt(1-x^2)

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

integral from-4 to x of (t-1)

\int_{- 4}^{x} (t - 1) d t

integral from 0 to ln(4) of e^{5x}

\int_{0}^{l n (4)} e^{5 x} d x