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人気のある微分積分 >

integral from-1 to 1 of 5/(1+25x^2)

解

\int_{- 1}^{1} \frac{5}{1 + 25 x ^{2}} d x

解

2 arctan (5)

+1

十進法表記

2.74680 \dots

解答ステップ

\int_{- 1}^{1} \frac{5}{1 + 25 x ^{2}} d x

定数を除く:

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

= 5 \cdot \int_{- 1}^{1} \frac{1}{1 + 25 x ^{2}} d x

置換積分法を適用する

= 5 \cdot \int_{- 5}^{5} \frac{1}{5 ( 1 + u ^{2} )} d u

定数を除く:

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

= 5 \cdot \frac{1}{5} \cdot \int_{- 5}^{5} \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)

= 5 \cdot \frac{1}{5} [arctan (u)]_{- 5}^{5}

簡素化

5 \cdot \frac{1}{5} [arctan (u)]_{- 5}^{5} : [arctan (u)]_{- 5}^{5}

= [arctan (u)]_{- 5}^{5}

限度を計算する:

2 arctan (5)

= 2 arctan (5)

グラフ

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