integral from 1/2 to 3/2 of qsqrt(q-1/2)

解

\int_{\frac{1}{2}}^{\frac{3}{2}} q q - \frac{1}{2} d q

\frac{11}{15}

十進法表記

0.73333 \dots

解答ステップ

\int_{\frac{1}{2}}^{\frac{3}{2}} q q - \frac{1}{2} d q

部分積分を適用する

= [\frac{1}{2 2} (q^{2} 2 q - 1 - 22 \cdot \int \frac{q ^{2}}{2 2 2 q - 1} d q)]_{\frac{1}{2}}^{\frac{3}{2}}

\int \frac{q ^{2}}{2 2 2 q - 1} d q = \frac{1}{30 2} (3 q^{2} + 2 q + 2) 2 q - 1

= [\frac{1}{2 2} (q^{2} 2 q - 1 - 22 \frac{1}{30 2} (3 q^{2} + 2 q + 2) 2 q - 1)]_{\frac{1}{2}}^{\frac{3}{2}}

簡素化

[\frac{1}{2 2} (q^{2} 2 q - 1 - 22 \frac{1}{30 2} (3 q^{2} + 2 q + 2) 2 q - 1)]_{\frac{1}{2}}^{\frac{3}{2}} : [\frac{1}{15 2} (6 q^{2} - q - 1) 2 q - 1]_{\frac{1}{2}}^{\frac{3}{2}}

= [\frac{1}{15 2} (6 q^{2} - q - 1) 2 q - 1]_{\frac{1}{2}}^{\frac{3}{2}}

限度を計算する:

\frac{11}{15}

= \frac{11}{15}