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人気のある微分積分 >

integral from 0 to 9 of (sqrt(x))/(x+9)

解

\int_{0}^{9} \frac{x}{x + 9} d x

解

6 (- \frac{π}{4} + 1)

+1

十進法表記

1.28761 \dots

解答ステップ

\int_{0}^{9} \frac{x}{x + 9} d x

u置換積分法を適用する

= \int_{0}^{3} \frac{2 u ^{2}}{u ^{2} + 9} d u

定数を除く:

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

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

\frac{u ^{2}}{u ^{2} + 9} = - \frac{9}{u ^{2} + 9} + 1

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

置換積分法を適用する

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

定数を除く:

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

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

\int_{0}^{1} \frac{1}{v ^{2} + 1} d v = \frac{π}{4}

\int_{0}^{1} 1 d v = 1

= 2 \cdot 3 (- \frac{π}{4} + 1)

簡素化

= 6 (- \frac{π}{4} + 1)

グラフ

人気の例

integral from 0 to x of sqrt(t+3)

\int_{0}^{x} t + 3 d t

integral from 5 to 25 of 1/(x(ln(x))^8)

\int_{5}^{25} \frac{1}{x ( ln ( x ) ) ^{8}} d x

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

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

integral from 0 to 1 of zsqrt(1-z^2)

\int_{0}^{1} z 1 - z^{2} d z

integral from-7 to 3 of-x^2-3x

\int_{- 7}^{3} - x^{2} - 3 x d x