What is the integral from y to 2 of x^3sin(x) ?

The integral from y to 2 of x^3sin(x) is 4cos(2)+6sin(2)+y^3cos(y)-3y^2sin(y)-6ycos(y)+6sin(y)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\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)

Solution steps

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

Apply Integration By Parts

= [- 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}

Simplify

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

Compute the boundaries:

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)

What is the integral from y to 2 of x^3sin(x) ?
The integral from y to 2 of x^3sin(x) is 4cos(2)+6sin(2)+y^3cos(y)-3y^2sin(y)-6ycos(y)+6sin(y)