What is #int_(0)^(10) x|sinx|dx#?

#12pi + 10cos10 - sin10#,

which is about #29.852#.

#x|sinx|#:

graph{x|sinx| [-0.1, 10.05, -2.92, 8]}

This may be a strange approach, but you can find all the intersections of #xsinx# with the #x# axis and break this up into chunks... Then, we can take the negative sign of the chunks with negative area.

#xsinx = 0#

Barring #x = 0#, we have:

#=> x = pi, 2pi, 3pi# that are in #[0,10]#.

That means we have the following intervals to consider:

#[0,pi]#: positive integral value

#[pi,2pi]#: negative integral value

#[2pi,3pi]#: positive integral value

#[3pi,10]#: negative integral value

#xsinx#:

graph{xsinx [-0.1, 10.05, -8, 8]}

Hence:

#int_(0)^(10) x|sinx|dx#

#= int_(0)^(pi) xsinx - int_(pi)^(2pi) xsinx + int_(2pi)^(3pi) xsinx - int_(3pi)^(10) xsinxdx#

Do integration by parts once on one of these.

#int x sinxdx#

Let #u = x#, #du = dx#, #dv = sinxdx#, and #v = -cosx#. Then:

#=> -xcosx + int cosxdx#

#= -xcosx + sinx#

Therefore:

#color(blue)(int_(0)^(10) x |sinx|dx)#

#= |[-xcosx + sinx]|_(0)^(pi)#

#- |[-xcosx + sinx]|_(pi)^(2pi)#

#+ |[-xcosx + sinx]|_(2pi)^(3pi)#

#- |[-xcosx + sinx]|_(3pi)^(10)#

#= ([cancel(-)picancel(cospi) + cancel(sinpi)^(0)] - [cancel(-0cos0)^(0) + cancel(sin0)^(0)]) - ([-2picancel(cos2pi) + cancel(sin2pi)^(0)] - [cancel(-)picancel(cospi) + cancel(sinpi)^(0)]) + ([cancel(-)3picancel(cos3pi) + cancel(sin3pi)^(0)] - [-2picancel(cos2pi) + cancel(sin2pi)^(0)]) - ([-10cos10 + sin10] - [cancel(-)3picancel(cos3pi) + cancel(sin3pi)^(0)])#

#= pi + 2pi + pi + 3pi + 2pi + 10cos10 - sin10 + 3pi#

#=# #color(blue)(12pi + 10cos10 - sin10)#