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人気のある微分積分 >

integral from 9 to 16 of (sqrt(x))/(x-4)

解

\int_{9}^{16} \frac{x}{x - 4} d x

解

- 4 (\frac{ln ( 3 ) - ln ( 5 )}{2} - \frac{1}{2})

+1

十進法表記

3.02165 \dots

解答ステップ

\int_{9}^{16} \frac{x}{x - 4} d x

u置換積分法を適用する

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

定数を除く:

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

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

因数

u^{2} - 4 : - (- u^{2} + 4)

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

定数を除く:

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

= 2 (- \int_{3}^{4} \frac{u ^{2}}{- u ^{2} + 4} d u)

置換積分法を適用する

= 2 (- \int_{\frac{3}{2}}^{2} \frac{2 v ^{2}}{- v ^{2} + 1} d v)

定数を除く:

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

= 2 (- 2 \cdot \int_{\frac{3}{2}}^{2} \frac{v ^{2}}{- v ^{2} + 1} d v)

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= 2 (- 2 \cdot \int_{\frac{3}{2}}^{2} \frac{1}{- v ^{2} + 1} - 1 d v)

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

= 2 (- 2 (\int_{\frac{3}{2}}^{2} \frac{1}{- v ^{2} + 1} d v - \int_{\frac{3}{2}}^{2} 1 d v))

\int_{\frac{3}{2}}^{2} \frac{1}{- v ^{2} + 1} d v = \frac{ln ( 3 ) - ln ( 5 )}{2}

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

= 2 (- 2 (\frac{ln ( 3 ) - ln ( 5 )}{2} - \frac{1}{2}))

簡素化

= - 4 (\frac{ln ( 3 ) - ln ( 5 )}{2} - \frac{1}{2})

グラフ

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