integral from y to 2 of t^3sin(t)

解

\int_{y}^{2} t^{3} sin (t) d t

4 cos (2) + 6 sin (2) + y^{3} cos (y) - 3 y^{2} sin (y) - 6 y cos (y) + 6 sin (y)

解答ステップ

\int_{y}^{2} t^{3} sin (t) d t

部分積分を適用する

= [- t^{3} cos (t) - \int - 3 t^{2} cos (t) d t]_{y}^{2}

\int - 3 t^{2} cos (t) d t = - 3 (t^{2} sin (t) - 2 (- t cos (t) + sin (t)))

= [- t^{3} cos (t) - (- 3 (t^{2} sin (t) - 2 (- t cos (t) + sin (t))))]_{y}^{2}

簡素化

= [- t^{3} cos (t) + 3 (t^{2} sin (t) - 2 (- t cos (t) + sin (t)))]_{y}^{2}

限度を計算する:

4 cos (2) + 6 sin (2) + y^{3} cos (y) - 3 y^{2} sin (y) - 6 y cos (y) + 6 sin (y)

= 4 cos (2) + 6 sin (2) + y^{3} cos (y) - 3 y^{2} sin (y) - 6 y cos (y) + 6 sin (y)