Integration by parts problem, college calc II?

Question

For what "c" is -1/2 the average value of (x-c)sin(x) over the interval [0,pi/3]? Thanks!

Answer 1

Average = 1/(π/3 - 0)[∫xsinx dx - ∫csinxdx] over [0,π/3]

∫xsinx dx over [0,π/3]    (let x = u and sinx = dv)

= x(-cosx) over [0,π/3] - ∫1.(-cosx)dx over [0,π/3]

= π/3(-1/2) + sinx over [0,π/3]

= -π/6 + √(3)/2

∫csinxdx] over [0,π/3] = -ccosx over [0,π/3]

= -c - (-c/2) = -c/2

Hence the avg = 1/(π/3) (-π/6 + √(3)/2 + c/2) = -1/2

Now solve for c after you've checked my work, because I haven't.