In my text book it shows that i should have the curves equal to y. So that I can find the x intercepts and work with the integral from there. However I can seem to figure out a solution for y^2=16x and 4x-3y=4. When I tried i got y=4 sqrt x and y=-(4(-x-1)/3) but i dont think thats right.
I suggest finding the area between x = y^2/16 and x = (4 + 3y)/4, in which case you need the y-coordinates of the intersection points of the two curves.
Equate: y^2/16 = (4 + 3y)/4. Solve for y.
Your values of y are correct (to 4dp - but are you meant to use exact values?)
Using your values of y as the integral terminals, I don't get your answer (but since you only give a final answer, it's impossible to know whether your error is in the calculus, algebra or arithmetic).