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人気のある微分積分 >

integral from 0 to 1 of (5/(7+7x^2))

解

\int_{0}^{1} (\frac{5}{7 + 7 x ^{2}}) d x

解

\frac{5 π}{28}

+1

十進法表記

0.56099 \dots

解答ステップ

\int_{0}^{1} \frac{5}{7 + 7 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_{0}^{1} \frac{1}{7 + 7 x ^{2}} d x

\frac{1}{7 + 7 x ^{2}} = \frac{1}{7} \frac{1}{1 + x ^{2}}

= 5 \cdot \int_{0}^{1} \frac{1}{7} \cdot \frac{1}{1 + 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 \frac{1}{7} \cdot \int_{0}^{1} \frac{1}{1 + x ^{2}} d x

共通積分を使用する:

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

= 5 \cdot \frac{1}{7} [arctan (x)]_{0}^{1}

簡素化

5 \cdot \frac{1}{7} [arctan (x)]_{0}^{1} : \frac{5}{7} [arctan (x)]_{0}^{1}

= \frac{5}{7} [arctan (x)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= \frac{5}{7} \cdot \frac{π}{4}

簡素化

= \frac{5 π}{28}

グラフ

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