integral from 0 to 4 of (2x+1)^{1/2}

解

\int_{0}^{4} (2 x + 1)^{\frac{1}{2}} d x

\frac{26}{3}

十進法表記

8.66666 \dots

解答ステップ

\int_{0}^{4} (2 x + 1)^{\frac{1}{2}} d x

部分積分を適用する

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

\int \frac{x}{( 2 x + 1 ) ^{\frac{1}{2}}} d x = \frac{1}{3} (x - 1) (2 x + 1)^{\frac{1}{2}}

= [x (2 x + 1)^{\frac{1}{2}} - \frac{1}{3} (x - 1) (2 x + 1)^{\frac{1}{2}}]_{0}^{4}

簡素化

[x (2 x + 1)^{\frac{1}{2}} - \frac{1}{3} (x - 1) (2 x + 1)^{\frac{1}{2}}]_{0}^{4} : [\frac{1}{3} (2 x (2 x + 1)^{\frac{1}{2}} + (2 x + 1)^{\frac{1}{2}})]_{0}^{4}

= [\frac{1}{3} (2 x (2 x + 1)^{\frac{1}{2}} + (2 x + 1)^{\frac{1}{2}})]_{0}^{4}

限度を計算する:

\frac{26}{3}

= \frac{26}{3}