If I understand your notation then
INT(0to1) 4y^3 dy = y^4 evaluated 1,0
int (0 to 1) 4y^3= 1
Thus the area under the curve is 1
int(0 to 1/2) 4y^3 dy = y^4 evaluated (1/2,0)
int(0 to 1/2) 4y^3= [1/2]^4 = 1/16
P(0<= y <= 1/2) = [1/16]/1
P(0<= y <= 1/2)=1/16 answer
I believe this is what your instructor wants
First find the area beneath the curve over its domain, and then the area beneath the curve over the domain of interest.
Arthur
Arthur