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