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人気のある微分積分 >

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

解

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

解

\frac{1}{3} (arctan (\frac{4}{3}) - arctan (\frac{2}{3}))

+1

十進法表記

0.11309 \dots

解答ステップ

\int_{1}^{2} \frac{2}{9 + 4 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_{1}^{2} \frac{1}{9 + 4 x ^{2}} d x

置換積分法を適用する

= 2 \cdot \int_{\frac{2}{3}}^{\frac{4}{3}} \frac{1}{6 ( u ^{2} + 1 )} 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}{6} \cdot \int_{\frac{2}{3}}^{\frac{4}{3}} \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)

= 2 \cdot \frac{1}{6} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

簡素化

2 \cdot \frac{1}{6} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}} : \frac{1}{3} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

= \frac{1}{3} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

限度を計算する:

arctan (\frac{4}{3}) - arctan (\frac{2}{3})

= \frac{1}{3} (arctan (\frac{4}{3}) - arctan (\frac{2}{3}))

グラフ

人気の例

integral from 1 to 3 of 8

\int_{1}^{3} 8 d x

integral from 0 to pi/2 of cos(2x-pi/2)

\int_{0}^{\frac{π}{2}} cos (2 x - \frac{π}{2}) d x

integral from 0 to 1 of r

\int_{0}^{1} r d r

integral from 0 to 4 of sqrt(1+4x^2)

\int_{0}^{4} 1 + 4 x^{2} d x

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

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