Evaluate the integral integrate from 1 to 2 of (x)sin(x/2)dx

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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This should be done by integration by parts.

Let;

`u = 2cos(x/2)`

`v = x`

`du = sin(x/2)dx`

`dv = 1dx`

 

By integration by parts;

`intvdu = uv-intudv`

`int^2_1 xsin(x/2) dx` = `[x*2cos(x/2)]^2_1-int^2_12cosx/2dx`

                        = `[x*2cos(x/2)]^2_1-4[sin(x/2)]^2_1`

                        = (3.999-1.999)-0.035

                        = 1.965

 

`int^2_1 xsin(x/2) dx= 1.965`

Sources:

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