It really depends on what you mean by integral.
Consider the indefinite integral of the indefinite intgeral of a function #f#:
#int int f(x) dx dx#
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#.
