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

解

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

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} x^{3} sin (x) d x

部分積分を適用する

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

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

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

簡素化

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