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人気のある微分積分 >

integral from 0 to 1/3 of 2/(1+9x^2)

解

\int_{0}^{\frac{1}{3}} \frac{2}{1 + 9 x ^{2}} d x

解

\frac{π}{6}

+1

十進法表記

0.52359 \dots

解答ステップ

\int_{0}^{\frac{1}{3}} \frac{2}{1 + 9 x ^{2}} d x

定数を除く:

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

= 2 \cdot \int_{0}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

置換積分法を適用する

= 2 \cdot \int_{0}^{1} \frac{1}{3 ( 1 + u ^{2} )} d u

定数を除く:

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

= 2 \cdot \frac{1}{3} \cdot \int_{0}^{1} \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)

= 2 \cdot \frac{1}{3} [arctan (u)]_{0}^{1}

簡素化

2 \cdot \frac{1}{3} [arctan (u)]_{0}^{1} : \frac{2}{3} [arctan (u)]_{0}^{1}

= \frac{2}{3} [arctan (u)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= \frac{2}{3} \cdot \frac{π}{4}

簡素化

= \frac{π}{6}

グラフ

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