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人気のある微分積分 >

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

解

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

解

\frac{1}{2} (arctan (2 \cdot 2) - arctan (2))

+1

十進法表記

0.19490 \dots

解答ステップ

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

置換積分法を適用する

= \int_{2}^{2 \cdot 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

= \frac{1}{2} \cdot \int_{2}^{2 \cdot 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)

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

限度を計算する:

arctan (2 \cdot 2) - arctan (2)

= \frac{1}{2} (arctan (2 \cdot 2) - arctan (2))

グラフ

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