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人気のある微分積分 >

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

解

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

解

\frac{1}{2}

+1

十進法表記

0.5

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

= 9 \cdot \int_{1}^{3} \frac{u - 1}{4 u ^{3}} d u

定数を除く:

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

= 9 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{u - 1}{u ^{3}} d u

拡張

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

= 9 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{1}{u ^{2}} - \frac{1}{u ^{3}} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 \cdot \frac{1}{4} (\int_{1}^{3} \frac{1}{u ^{2}} d u - \int_{1}^{3} \frac{1}{u ^{3}} d u)

\int_{1}^{3} \frac{1}{u ^{2}} d u = \frac{2}{3}

\int_{1}^{3} \frac{1}{u ^{3}} d u = \frac{4}{9}

= 9 \cdot \frac{1}{4} (\frac{2}{3} - \frac{4}{9})

簡素化

9 \cdot \frac{1}{4} (\frac{2}{3} - \frac{4}{9}) : \frac{1}{2}

= \frac{1}{2}

グラフ

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