Let f(x) be defined and continuous in [a,b] and g(x) defined and differantiable in [c,d] with values in [a,b], such that g(c) =a and g(d) = b.
Suppose also for simplicity that g'(x) >0.
The chain rule states that:
d/dx [f(g(x))] = f'(g(x)) g'(x)
Consider now the definite integral as limit of the Riemann sum:
int_c^d f'(g(x)) g'(x)dx = lim_(N->oo) sum_(n=0)^N f(g(xi_k))g'(xi_k)(x_(k+1)-x_k)
where x_0 < x_1 < ... < x_N in [c,d] and xi_k in (x_k,x_(k+1)).
As the choice of xi_k in the interval is arbitrary we can choose them as the points for which, based on Lagrange's theorem:
g'(xi_k) = (g(x_(k+1))-g(x_k))/(x_(k+1)-x_k)
so:
int_c^d f'(g(x)) g'(x)dx = lim_(N->oo) sum_(n=0)^N f'(g(xi_k))(g(x_(k+1))-g(x_k))
Let now: y_k = g(x_k). As g'(x) > 0 the function is strictly increasing so: y_0 < y_1 <... < y_N
Then:
int_a^b f'(g(x)) g'(x)dx = lim_(N->oo) sum_(n=0)^N f'(y_k)(y_(k+1)-y_k)
but this is a Riemann sum of f(y) for y in [a,b], so:
int_c^d f'(g(x)) g'(x)dx = int_a^b f'(y) dy = f(b)-f(a)
