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人気のある微分積分 >

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

解

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

解

arctan (2) - \frac{π}{4}

+1

十進法表記

0.32175 \dots

解答ステップ

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

共通積分を使用する:

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

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

限度を計算する:

arctan (2) - \frac{π}{4}

= arctan (2) - \frac{π}{4}

グラフ

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