It really depends on what you mean by integral.
Consider the indefinite integral of the indefinite intgeral of a function ff:
int int f(x) dx dx∫∫f(x)dxdx
If f(x) = g^(prime)(x) = h^(prime prime) (x), that is, f is a double antiderivative, then, by applying the Fundamental Theorem of Calculus twice:
int int f(x) dx dx = int int g^(prime) (x) dx dx = int [ g(x) + C ]dx = int [ h^(prime) (x) + C ]dx = int h^(prime) (x) dx + int C dx = h (x) + Cx + D,
where C and D are arbitrary constants.
Now consider the definite integral of the definite integral of a function f = g^(prime)(x). Then, applying the Fundamental Theorem of Calculus again:
int_c^d int_a^b f(x) dx dx = int_c^d int_a^b g^(prime) (x) dx dx = int_c^d [g(b)-g(a)] dx
But g(b)-g(a) is simply a real number. For int_c^d [g(b)-g(a)] dx to make sense, we must consider g(b)-g(a) as a constant function.
Integrating again, we get:
int_c^d [g(b)-g(a)] dx = [g(b)-g(a)]x|_c^d = [g(b)-g(a)] (d-c)
A third interpretation could also be that of iterated integrals, that appear in multivariable calculus.
Consider the two variable function f(x,y). There can be two kinds of iterated integrals:
int_(y_1)^(y_2) [ int_(x_1)^(x_2) f(x,y) dx ] dy
and
int_(x_1)^(x_2) [ int_(y_1)^(y_2) f(x,y) dy ] dx
If f(x,y) = (partial)/(partial x) g_1(x,y), g_1 (x_1,y)= d/(dy)h_1 (y) and g_1 (x_2,y)= d/(dy)p_1 (y), then we have the following result:
int_(y_1)^(y_2) [ int_(x_1)^(x_2) f(x,y) dx ] dy = int_(y_1)^(y_2) [(g_1(x_2,y)-g_1(x_1,y)] dy = [p_1(y_2)-p_1(y_1)] - [h_1(y_2) - h_1(y_1)]
An analogous result holds for int_(x_1)^(x_2) [ int_(y_1)^(y_2) f(x,y) dy ] dx, although, in general,
int_(y_1)^(y_2) [ int_(x_1)^(x_2) f(x,y) dx ] dy ne int_(x_1)^(x_2) [ int_(y_1)^(y_2) f(x,y) dy ] dx.
