second integral mean-value theorem

If the real functions $f$ and $g$ are continuous and $f$ monotonic on the interval  $[a,b]$,  then the equation

${\displaystyle {\int }_a^b}f(x)g(x)𝑑x=f(a){\displaystyle {\int }_a^{\xi }}g(x)𝑑x+f(b){\displaystyle {\int }_{\xi }^b}g(x)𝑑x$(1)

is true for a value $\xi$ in this interval.

Proof.  We can suppose that  $f(a)\ne f(b)$  since otherwise any value of $\xi$ between $a$ and $b$ would do.

Let’s first prove the auxiliary result, that if a function $\phi$ is continuous on an open interval $I$ containing $[a,b]$  then

$\underset{h\to 0}{lim}{\displaystyle {\int }_a^b}{\displaystyle \frac{\phi (x+h)-\phi (x)}{h}}𝑑x=\phi (b)-\phi (a).$(2)

In fact, when we take an antiderivative $\mathrm{\Phi }$ of $\phi$, then for every nonzero $h$ the function

$$x\mapsto \frac{\mathrm{\Phi }(x+h)-\mathrm{\Phi }(x)}{h}$$

is an antiderivative of the integrand of (2) on the interval  $[a,b]$.  Thus we have

$${\int }_a^b\frac{\phi (x+h)-\phi (x)}{h}𝑑x=\frac{\mathrm{\Phi }(b+h)-\mathrm{\Phi }(x)}{h}-\frac{\mathrm{\Phi }(a+h)-\mathrm{\Phi }(x)}{h}\to \phi (b)-\phi (a)\mathit{\hspace{1em}}\text{as}h\to \mathrm{\hspace{0.17em}0}.$$

The given functions $f$ and $g$ can be extended on an open interval $I$ containing  $[a,b]$  such that they remain continuous and $f$ monotonic.  We take an antiderivative $G$ of $g$ and a nonzero number $h$ having small absolute value.  Then we can write the identity

${\displaystyle {\int }_a^b}{\displaystyle \frac{f(x+h)G(x+h)-f(x)G(x)}{h}}𝑑x={\displaystyle {\int }_a^b}{\displaystyle \frac{f(x+h)[G(x+h)-G(x)]}{h}}𝑑x-{\displaystyle {\int }_a^b}{\displaystyle \frac{f(x+h)-f(x)}{h}}G(x)𝑑x.$(3)

By (2), the left hand side of (3) may be written

${\displaystyle {\int }_a^b}{\displaystyle \frac{f(x+h)G(x+h)-f(x)G(x)}{h}}𝑑x=f(b)G(b)-f(a)G(a)+{\epsilon }_1(h)$(4)

where  ${\epsilon }_1(h)\to 0$  as  $h\to 0$.  Further, the function

$x\mapsto \{\begin{array}{cc}\frac{f(x+h)[G(x+h)-G(x)]}{h}\hspace{1em}\text{for}\hspace{1em}h\ne \mathrm{\hspace{0.33em}0}\hfill & \\ f(x)g(x)\hspace{1em}\hspace{1em}\text{for}\hspace{1em}\hspace{1em}h=\mathrm{\hspace{0.33em}0}\hfill & \end{array}$

is continuous in a rectangle  $a\le x\le b,-\delta \le h\le \delta$,  whence we have

${\displaystyle {\int }_a^b}{\displaystyle \frac{f(x+h)[G(x+h)-G(x)]}{h}}𝑑x={\displaystyle {\int }_a^b}f(x)g(x)𝑑x+{\epsilon }_2(h)$(5)

where  ${\epsilon }_2(h)\to 0$  as  $h\to 0$.  Because of the monotonicity of $f$, the expression$\frac{f(x+h)-f(x)}{h}$ does not change its sign when  $a\le x\le b$.  Then the usual integral mean value theorem guarantees for every $h$ (sufficiently near 0) a number ${\xi }_h$ of the interval  $[a,b]$  such that

$${\int }_a^b\frac{f(x+h)-f(x)}{h}G(x)𝑑x=G({\xi }_h){\int }_a^b\frac{f(x+h)-f(x)}{h}𝑑x,$$

and the auxiliary result (2) allows to write this as

${\displaystyle {\int }_a^b}{\displaystyle \frac{f(x+h)-f(x)}{h}}G(x)𝑑x=G({\xi }_h)[f(b)-f(a)+{\epsilon }_3(h)]$(6)

with  ${\epsilon }_3(h)\to 0$  as  $h\to 0$.  Now the equations (4), (5) and (6) imply

$f(b)G(b)-f(a)G(a)+{\epsilon }_1(h)={\displaystyle {\int }_a^b}f(x)g(x)𝑑x+{\epsilon }_2(h)+G({\xi }_h)[f(b)-f(a)+{\epsilon }_3(h)].$(7)

Because  $f(b)-f(a)\ne 0$,  the expression $G({\xi }_h)$ has a limit $L$ for  $h\to 0$.  By the continuity of$G$ there must be a number $\xi$ between $a$ and $b$ such that  $G(\xi )=L$.  Letting then $h$ tend to 0 we thus get the limiting equation

$$f(b)G(b)-f(a)G(a)={\int }_a^{b}f(x)g(x)𝑑x+G(\xi )[f(b)-f(a)],$$

which finally gives

$${\int }_a^{b}f(x)g(x)𝑑x=f(a)[G(\xi )-G(a)]+f(b)[G(b)-G(\xi )]=f(a){\int }_a^{\xi }g(x)𝑑x+f(b){\int }_{\xi }^{b}g(x)𝑑x$$

Q.E.D.