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人気のある微分積分 >

integral from 1 to 3 of (2^x)/(2^x+4)

解

\int_{1}^{3} \frac{2 ^{x}}{2 ^{x} + 4} d x

解

1

解答ステップ

\int_{1}^{3} \frac{2 ^{x}}{2 ^{x} + 4} d x

u置換積分法を適用する

= \int_{6}^{12} \frac{1}{ln ( 2 ) u} d u

定数を除く:

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

= \frac{1}{ln ( 2 )} \cdot \int_{6}^{12} \frac{1}{u} d u

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{ln ( 2 )} [ln ∣ u ∣]_{6}^{12}

限度を計算する:

ln (12) - ln (6)

= \frac{1}{ln ( 2 )} (ln (12) - ln (6))

簡素化

= 1

グラフ

人気の例

integral from 0 to 3 of 3+2y-y^2

\int_{0}^{3} 3 + 2 y - y^{2} d y

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

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

integral from 0 to 1 of (9x^e+4e^x)

\int_{0}^{1} (9 x^{e} + 4 e^{x}) d x

integral from 0 to 8 of sqrt(2x)-(x-4)

\int_{0}^{8} 2 x - (x - 4) d x

integral from 0 to 8 of |x^2-6x+8|

\int_{0}^{8} x^{2} - 6 x + 8 d x