ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 0 to 2 of 4/(x^2+4)

解

\int_{0}^{2} \frac{4}{x ^{2} + 4} d x

解

\frac{π}{2}

+1

十進法表記

1.57079 \dots

解答ステップ

\int_{0}^{2} \frac{4}{x ^{2} + 4} d x

定数を除く:

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

= 4 \cdot \int_{0}^{2} \frac{1}{x ^{2} + 4} d x

置換積分法を適用する

= 4 \cdot \int_{0}^{1} \frac{1}{2 ( u ^{2} + 1 )} d u

定数を除く:

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

= 4 \cdot \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

共通積分を使用する:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 4 \cdot \frac{1}{2} [arctan (u)]_{0}^{1}

簡素化

4 \cdot \frac{1}{2} [arctan (u)]_{0}^{1} : 2 [arctan (u)]_{0}^{1}

= 2 [arctan (u)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= 2 \cdot \frac{π}{4}

簡素化

= \frac{π}{2}

グラフ

人気の例

integral from 0 to 2pi of cos(mx)cos(nx)

\int_{0}^{2 π} cos (m x) cos (n x) d x

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

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

integral from 0 to 4 of 2x(x^2+4)^2

\int_{0}^{4} 2 x (x^{2} + 4)^{2} d x

integral from 0 to 3 of 1/(16+x^2)

\int_{0}^{3} \frac{1}{16 + x ^{2}} d x

integral from-1 to 0 of (6x)

\int_{- 1}^{0} (6 x) d x