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人気のある微分積分 >

integral from 0 to 4/3 of 1/(9x^2+16)

解

\int_{0}^{\frac{4}{3}} \frac{1}{9 x ^{2} + 16} d x

解

\frac{π}{48}

+1

十進法表記

0.06544 \dots

解答ステップ

\int_{0}^{\frac{4}{3}} \frac{1}{9 x ^{2} + 16} d x

置換積分法を適用する

= \int_{0}^{1} \frac{1}{12 ( u ^{2} + 1 )} d u

定数を除く:

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

= \frac{1}{12} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

共通積分を使用する:

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

= \frac{1}{12} [arctan (u)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= \frac{1}{12} \cdot \frac{π}{4}

簡素化

= \frac{π}{48}

グラフ

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