I think the right question is  

f(x)  = ₀∫^x ( t²-t+2 )²⁰⁰⁵( t²-t-2 )²⁰⁰⁷( t²-t-6 )²⁰⁰⁹( t²-t-12 )²⁰¹¹( t²-3t+2 )²⁰¹²  dt    

then sum of values of x,  where maxima of f(x) will be obtained is

for maxima f '(x) = 0

by using Newton Lebnitz formula we get

f ^'(x) =  (x²-x+2 )²⁰⁰⁵( x²-x-2 )²⁰⁰⁷( x²-x-6 )²⁰⁰⁹( x²-x-12 )²⁰¹¹( x²-3x+2 )²⁰¹²

comparing it with 0 and making factors we will get x = -1, 2, 3, -2, 4, -3, 1

now obtain the maxima point then sum these values