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人気のある微分積分 >

integral from 1 to 16 of 1/(1+sqrt(x))

解

\int_{1}^{16} \frac{1}{1 + x} d x

解

2 (3 - ln (5) + ln (2))

+1

十進法表記

4.16741 \dots

解答ステップ

\int_{1}^{16} \frac{1}{1 + x} d x

u置換積分法を適用する

= \int_{2}^{5} \frac{2 ( u - 1 )}{u} d u

定数を除く:

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

= 2 \cdot \int_{2}^{5} \frac{u - 1}{u} d u

拡張

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 2 \cdot \int_{2}^{5} 1 - \frac{1}{u} 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

= 2 (\int_{2}^{5} 1 d u - \int_{2}^{5} \frac{1}{u} d u)

\int_{2}^{5} 1 d u = 3

\int_{2}^{5} \frac{1}{u} d u = ln (5) - ln (2)

= 2 (3 - (ln (5) - ln (2)))

簡素化

= 2 (3 - ln (5) + ln (2))

グラフ

人気の例

integral from 2 to infinity of u*e^{-u}

\int_{2}^{\infty} u \cdot e^{- u} d u

integral from-2 to 1 of x^3-2x^2-5x+6

\int_{- 2}^{1} x^{3} - 2 x^{2} - 5 x + 6 d x

integral from 0 to 1 of e^{-x}cos(wx)

\int_{0}^{1} e^{- x} cos (w x) d x

integral from 0 to pi/2 of (e^x+cos(x))

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

integral from-infinity to e of 16e^{-x}

\int_{- \infty}^{e} 16 e^{- x} d x