Modify the proof given in class of the First Mean Value Theo

Modify the proof (given in class) of the First Mean Value Theorem for Integrals to prove the Second Mean Value Theorem for Integrals: Let a and b be real numbers with a 0 for every a lessthanorequalto x lessthanorequalto b. Then there is a c in [a, b] such that integral_a^b f(x)g(x)dx = f(c) integral_a^b g(x) dx. (The hypotheses on g guarantee that integral_a^b g(x) dx > 0, in particular that the said integral is nonzero. This may be taken for granted. Observe that the First Mean Value Theorem follows from the Second by choosing g(x) = 1 for every a lessthanorequalto x lessthanorequalto b.)

Solution

Let ff and gg be defined on [a,b][a,b] with gg continuous, f0f0, and ff integrable. Then there is a point x0(a,b)x0(a,b) such that
baf(x)g(x)dx=g(x0)baf(x)dx.
abf(x)g(x)dx=g(x0)abf(x)dx.
My (incomplete) proof goes as follows:

Since [a,b][a,b] is compact and gg is continuous, we know there exist x1,x2[a,b]x1,x2[a,b] such that g(x1)g(x)g(x2)g(x1)g(x)g(x2) for any x[a,b]x[a,b]. Thus, we have the following string of inequalities
f(x)g(x1)f(x)g(x)f(x)g(x2),x[a,b],
f(x)g(x1)f(x)g(x)f(x)g(x2),x[a,b],
g(x1)baf(x)dxbaf(x)g(x)dxg(x2)baf(x)dx.
g(x1)abf(x)dxabf(x)g(x)dxg(x2)abf(x)dx.

Let h(x)=g(x)baf(t) dth(x)=g(x)abf(t) dt. As gg is continuous, hh is also continuous. Without loss of generality, let x1<x2x1<x2.

By what you\'ve shown above, baf(x)g(x) dxabf(x)g(x) dx is a number between h(x1)h(x1) and h(x2)h(x2). As hh is continuous, by the IVP there must be a value x0(x1,x2)x0(x1,x2) such that

h(x0)=baf(x)g(x) dxh(x0)=abf(x)g(x) dx

That is, there is an x0(x1,x2)(a,b) x0(x1,x2)(a,b)  such that

g(x0)baf(x) dx=baf(x)g(x) dx

Let h(x)=g(x)baf(t) dth(x)=g(x)abf(t) dt. As gg is continuous, hh is also continuous. Without loss of generality, let x1<x2x1<x2. By what you\'ve shown above, baf(x)g(x) dxabf(x)g(x) dx is a number between h(x1)h(x1) and h(x2)h(x2). As hh is continuous, by the IVP there must be a value x0(x1,x2)x0(x1,x2) such that h(x0)=baf(x)g(x) dxh(x0)=abf(x)g(x) dx That is, there is an x0(x1,x2)(a,b) x0(x1,x2)(a,b)  such that g(x0)baf(x) dx=baf(x)g(x) dx