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人気のある微分積分 >

integral from-1 to 1 of 2/(1+4t^2)

解

\int_{- 1}^{1} \frac{2}{1 + 4 t ^{2}} d t

解

2 arctan (2)

+1

十進法表記

2.21429 \dots

解答ステップ

\int_{- 1}^{1} \frac{2}{1 + 4 t ^{2}} d t

定数を除く:

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

= 2 \cdot \int_{- 1}^{1} \frac{1}{1 + 4 t ^{2}} d t

置換積分法を適用する

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

定数を除く:

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

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

共通積分を使用する:

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

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

簡素化

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

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

限度を計算する:

2 arctan (2)

= 2 arctan (2)

グラフ

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