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人気のある微分積分 >

integral from 0 to 2 of x*arctan(x/2)

解

\int_{0}^{2} x \cdot arctan (\frac{x}{2}) d x

解

π - 2

+1

十進法表記

1.14159 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{0}^{1} 4 u arctan (u) d u

定数を除く:

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

= 4 \cdot \int_{0}^{1} u arctan (u) d u

部分積分を適用する

= 4 [\frac{1}{2} u^{2} arctan (u) - \int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u]_{0}^{1}

\int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} (- arctan (u) + u)

= 4 [\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u)]_{0}^{1}

簡素化

4 [\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u)]_{0}^{1} : 4 [\frac{1}{2} (u^{2} arctan (u) + arctan (u) - u)]_{0}^{1}

= 4 [\frac{1}{2} (u^{2} arctan (u) + arctan (u) - u)]_{0}^{1}

限度を計算する:

\frac{π - 2}{4}

= 4 \cdot \frac{π - 2}{4}

簡素化

= π - 2

グラフ

人気の例

integral from 0 to 1/4 of 4sqrt(1-16x^2)

\int_{0}^{\frac{1}{4}} 4 1 - 16 x^{2} d x

integral from 1 to 2 of (x/y-y/x)

\int_{1}^{2} (\frac{x}{y} - \frac{y}{x}) d y

integral from 0 to 1 of h(x-x^2)

\int_{0}^{1} h (x - x^{2}) d x

integral from 2x to x^2 of sqrt(t^2+1)

\int_{2 x}^{x^{2}} t^{2} + 1 d t

integral from 0 to 1/3 of xln(3x)

\int_{0}^{\frac{1}{3}} x ln (3 x) d x