integral from-1 to 0 of t/((3t+4)^3)

解

\int_{- 1}^{0} \frac{t}{( 3 t + 4 ) ^{3}} d t

- \frac{1}{8}

十進法表記

- 0.125

解答ステップ

\int_{- 1}^{0} \frac{t}{( 3 t + 4 ) ^{3}} d t

部分積分を適用する

= [- \frac{t}{6 ( 3 t + 4 ) ^{2}} - \int - \frac{1}{6 ( 3 t + 4 ) ^{2}} d t]_{- 1}^{0}

\int - \frac{1}{6 ( 3 t + 4 ) ^{2}} d t = \frac{1}{18 ( 3 t + 4 )}

= [- \frac{t}{6 ( 3 t + 4 ) ^{2}} - \frac{1}{18 ( 3 t + 4 )}]_{- 1}^{0}

結合

- \frac{t}{6 ( 3 t + 4 ) ^{2}} - \frac{1}{18 ( 3 t + 4 )} : - \frac{3 t + 2}{9 ( 3 t + 4 ) ^{2}}

= [- \frac{3 t + 2}{9 ( 3 t + 4 ) ^{2}}]_{- 1}^{0}

限度を計算する:

- \frac{1}{8}

= - \frac{1}{8}