integral from 1 to 2 of tsqrt(1/2 t^2+9)

解

\int_{1}^{2} t \frac{1}{2} t^{2} + 9 d t

\frac{22 11}{3} - \frac{19 19}{3 2}

十進法表記

4.80127 \dots

解答ステップ

\int_{1}^{2} t \frac{1}{2} t^{2} + 9 d t

部分積分を適用する

= [\frac{1}{2 2} t^{2} t^{2} + 18 - \int \frac{t ^{3}}{2 2 t ^{2} + 18} d t]_{1}^{2}

\int \frac{t ^{3}}{2 2 t ^{2} + 18} d t = \frac{1}{2 2} (\frac{1}{3} (t^{2} + 18)^{\frac{3}{2}} - 18 t^{2} + 18)

= [\frac{1}{2 2} t^{2} t^{2} + 18 - \frac{1}{2 2} (\frac{1}{3} (t^{2} + 18)^{\frac{3}{2}} - 18 t^{2} + 18)]_{1}^{2}

簡素化

[\frac{1}{2 2} t^{2} t^{2} + 18 - \frac{1}{2 2} (\frac{1}{3} (t^{2} + 18)^{\frac{3}{2}} - 18 t^{2} + 18)]_{1}^{2} : [\frac{1}{3 2} (t^{2} + 18) t^{2} + 18]_{1}^{2}

= [\frac{1}{3 2} (t^{2} + 18) t^{2} + 18]_{1}^{2}

限度を計算する:

\frac{22 11}{3} - \frac{19 19}{3 2}

= \frac{22 11}{3} - \frac{19 19}{3 2}