integral from 1/2 to e/2 of tln(2t)

解

\int_{\frac{1}{2}}^{\frac{e}{2}} t ln (2 t) d t

\frac{e ^{2} + 1}{16}

十進法表記

0.52431 \dots

解答ステップ

\int_{\frac{1}{2}}^{\frac{e}{2}} t ln (2 t) d t

部分積分を適用する

= [\frac{1}{2} (t^{2} ln (2 t) - 2 \cdot \int \frac{t}{2} d t)]_{\frac{1}{2}}^{\frac{e}{2}}

\int \frac{t}{2} d t = \frac{t ^{2}}{4}

= [\frac{1}{2} (t^{2} ln (2 t) - 2 \cdot \frac{t ^{2}}{4})]_{\frac{1}{2}}^{\frac{e}{2}}

2 \cdot \frac{t ^{2}}{4} = \frac{t ^{2}}{2}

= [\frac{1}{2} (t^{2} ln (2 t) - \frac{t ^{2}}{2})]_{\frac{1}{2}}^{\frac{e}{2}}

限度を計算する:

\frac{e ^{2}}{16} + \frac{1}{16}

= \frac{e ^{2}}{16} + \frac{1}{16}

簡素化

= \frac{e ^{2} + 1}{16}