I know how to do PART A I am having difficulties with part B finding the area!

Part A:

-5/6 t³ + 5t + 1 = -5/6 [-t³] + 5[-t] + 1

-5/6 t³ + 5t = 0

t³ - 6t = 0

t = 0

t = +/-√6

For the Area

S = integral of y dx

S = integral of [-5/6 t³ + 5t + 1] d[6 - 5/2 t²]

S = integral of [-5/6 t³ + 5t + 1] 5tdt from T+ to T-

But I am not getting the right answer... Can anyone help?