Modify the proof (given in class) of the First Mean Value Theorem for Integrals
ID: 3405176 • Letter: M
Question
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.)Explanation / Answer
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
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