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人気のある微分積分 >

integral from 0 to 1 of (1-e^2)^{1/2}

解

\int_{0}^{1} (1 - e^{2})^{\frac{1}{2}} d x

解

i (e^{2} - 1)^{\frac{1}{2}}

解答ステップ

\int_{0}^{1} (1 - e^{2})^{\frac{1}{2}} d x

定数の積分:

\int a d x = a x

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

虚数の規則を適用する:

- a = i a

= [i (- 1 + e^{2})^{\frac{1}{2}} x]_{0}^{1}

限度を計算する:

i (e^{2} - 1)^{\frac{1}{2}}

= i (e^{2} - 1)^{\frac{1}{2}}

人気の例

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

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

integral from-pi to pi of (sin(x))^2

\int_{- π}^{π} (sin (x))^{2} d x

integral from 0 to 1 of sqrt(1+2t^2+t^4)

\int_{0}^{1} 1 + 2 t^{2} + t^{4} d t

integral from e to e^2 of (ln(x))^2(1/x)

\int_{e}^{e^{2}} (ln (x))^{2} (\frac{1}{x}) d x

integral from-1 to 1 of (x+1)/(x-1)

\int_{- 1}^{1} \frac{x + 1}{x - 1} d x