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人気のある微分積分 >

integral from 1 to 2 of e^x+1/x

解

\int_{1}^{2} e^{x} + \frac{1}{x} d x

解

e^{2} - e + ln (2)

+1

十進法表記

5.36392 \dots

解答ステップ

\int_{1}^{2} e^{x} + \frac{1}{x} d x

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

= \int_{1}^{2} e^{x} d x + \int_{1}^{2} \frac{1}{x} d x

\int_{1}^{2} e^{x} d x = e^{2} - e

\int_{1}^{2} \frac{1}{x} d x = ln (2)

= e^{2} - e + ln (2)

グラフ

人気の例

integral from 0 to 1 of e^{2x-1}

\int_{0}^{1} e^{2 x - 1} d x

integral from 0 to u of 1/x

\int_{0}^{u} \frac{1}{x} d x

integral from-4 to 4 of-5

\int_{- 4}^{4} - 5 d x

integral from 2 to 4 of 4x^2-5x+8

\int_{2}^{4} 4 x^{2} - 5 x + 8 d x

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

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