I was playing around with some integrals and I saw that:
the desired value is $a$ but can't it can be approximated by
$$f(a)=\int_{c_z}^a f'(x)dx=\int_{c_z}^b f'(x) dx+\int_b^c f'(x)dx+\int_c^df'(x) dx.......$$ they follow the pattern $$c_z < b < c < d<......< a$$
Is this method valid? Note: $f(c_z)=0$
I would say yes. The indefinite integral of the derivative of a function is just the function (when a constant is added), and the definite integral of the derivative of a function is just the value of the function (provided the lower bound is where the function is equal to $0$ ).
Some examples: $$f(x) = x^2 - 1,\;\; f'(x) = 2x,\;\; \int_1^32x\;dx = 8 = f(3)$$ $$g(x) = 3x+6, \;\; g'(x) = 3, \;\; \int_{-2}^23\;dx + \int_2^43\;dx = 12 +6 = g(4)$$ In fact, you don't even need the qualification $c_z < b < c < d \;. . . < a$ $$h(x) = x^2-x,\;\;h'(x) = 2x-1,\;\;\int_1^02x-1\;dx + \int_0^22x-1\;dx = 0+2=h(2)$$