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人気のある微分積分 >

integral from 0 to 1 of 6/(t^2+1)

解

\int_{0}^{1} \frac{6}{t ^{2} + 1} d t

解

\frac{3 π}{2}

+1

十進法表記

4.71238 \dots

解答ステップ

\int_{0}^{1} \frac{6}{t ^{2} + 1} d t

定数を除く:

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

= 6 \cdot \int_{0}^{1} \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)

= 6 [arctan (t)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= 6 \cdot \frac{π}{4}

簡素化

= \frac{3 π}{2}

グラフ

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