Answer:
f(1) ≈ 2.7864
Step-by-step explanation:
You appear to want a couple of iterations of ...
... y[n+1] = y[n] +arcsin(x[n]·y[n]}·(x[n+1] -x[n])
... x[n+1] = x[n] +0.5
... x[0] = 0
... y[0] = 2
Filling in the values, we get
... y[1] = 2 + arcsin(0·2)·0.5 = 2
... y[2] = 2 + arcsin(0.5·2)·0.5 = 2 +(π/2)·0.5 ≈ 2.7864 . . . . corresponds to x=1
