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人気のある微分積分 >

integral from 0 to ln(2x) of sin(5e^t)

解

\int_{0}^{l n (2 x)} sin (5 e^{t}) d t

解

Si (10 x) - 1.54993 \dots

解答ステップ

\int_{0}^{l n (2 x)} sin (5 e^{t}) d t

u置換積分法を適用する

= \int_{5}^{10 x} \frac{sin ( u )}{u} d u

非初等積分を使用する:

\int \frac{sin ( x )}{x} d x = Si (x)

= [Si (u)]_{5}^{10 x}

限度を計算する:

Si (10 x) - 1.54993 \dots

= Si (10 x) - 1.54993 \dots

人気の例

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

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

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

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

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

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

integral from 0 to pi of t^2

\int_{0}^{π} t^{2} d t

integral from 0 to 4 of xy

\int_{0}^{4} x y d y