Help solve the Riemann sum of -x^2/3-4 for [0,2]?

Sol
f(x)=－x^2/3－4
(0，2)分成n份
f(2k/n)=－(4k^2/n^2)/3－4=(－4/3)(k^2/n^2)－4
S(n)=Σ(k=1 to n)_[(－4/3)(k^2/n^2)－4]*(2/n)
=Σ(k=1 to n)_[(－4/3)(k^2/n^2)]*(2/n)－Σ(k=1 to n)_4*(2/n)
=[－8/(3n^3)]*Σ(k=1 to n)_k^2－(8/n)*Σ(k=1 to n)_1
=[－8/(3n^3)]*[n(n+1)(2n+1)/6]－(8/n)*n
=(－4/9)*(2n^3+3n^2+n)/n^3－8
=(－8/9)－(4/3)/n－(4/9)/n^2－8
=(－80/9)－(4/3)/n－(4/9)/n^2
S=lim(n->∞)_[(－80/9)－(4/3)/n－(4/9)/n^2]=－80/9
*******
∫(0 to 2)_(－x^2/3－4)dx
=－x^3/9－4x｜(0 to 2)
=[(－8/9)－8]－0
=－80/9