# 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

 $\displaystyle\int_{a}^{b}\!f(x)g(x)\,dx\;=\;f(a)\!\int_{a}^{\,\xi}\!g(x)\,dx+f% (b)\!\int_{\xi}^{\,b}\!g(x)\,dx$ (1)

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

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

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

 $\displaystyle\lim_{h\to 0}\int_{a}^{b}\!\frac{\varphi(x\!+\!h)\!-\!\varphi(x)}% {h}\,dx\;=\;\varphi(b)\!-\varphi(a).$ (2)

In fact, when we take an antiderivative $\Phi$ of $\varphi$, then for every nonzero $h$ the function

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

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

 $\int_{a}^{b}\!\frac{\varphi(x\!+\!h)\!-\!\varphi(x)}{h}\,dx\;=\;\frac{\Phi(b\!% +\!h)\!-\!\Phi(x)}{h}-\frac{\Phi(a\!+\!h)\!-\!\Phi(x)}{h}\;\to\;\varphi(b)\!-% \!\varphi(a)\quad\mbox{as}\;\;h\,\to\,0.$

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

 $\displaystyle\int_{a}^{b}\!\frac{f(x\!+\!h)G(x\!+\!h)\!-\!f(x)G(x)}{h}\,dx\;=% \;\int_{a}^{b}\!\frac{f(x\!+\!h)[G(x\!+\!h)\!-\!G(x)]}{h}\,dx-\int_{a}^{b}\!% \frac{f(x\!+\!h)\!-\!f(x)}{h}G(x)\,dx.$ (3)

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

 $\displaystyle\int_{a}^{b}\!\frac{f(x\!+\!h)G(x\!+\!h)\!-\!f(x)G(x)}{h}\,dx\;=% \;f(b)G(b)\!-\!f(a)G(a)\!+\!\varepsilon_{1}(h)$ (4)

where  $\varepsilon_{1}(h)\to 0$  as  $h\to 0$.  Further, the function

 $\displaystyle x\;\mapsto\;\begin{cases}\frac{f(x+h)[G(x+h)-G(x)]}{h}\quad\mbox% {for}\quad h\;\neq\;0\\ f(x)g(x)\qquad\mbox{for}\qquad h\;=\;0\end{cases}$

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

 $\displaystyle\int_{a}^{b}\!\frac{f(x\!+\!h)[G(x\!+\!h)\!-\!G(x)]}{h}\,dx\;=\;% \int_{a}^{b}\!f(x)g(x)\,dx+\varepsilon_{2}(h)$ (5)

where  $\varepsilon_{2}(h)\to 0$  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  $a\leq x\leq b$.  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

 $\displaystyle\int_{a}^{b}\!\frac{f(x\!+\!h)\!-\!f(x)}{h}G(x)\,dx\;=\;G(\xi_{h}% )[f(b)\!-\!f(a)\!+\!\varepsilon_{3}(h)]$ (6)

with  $\varepsilon_{3}(h)\to 0$  as  $h\to 0$.  Now the equations (4), (5) and (6) imply

 $\displaystyle f(b)G(b)\!-\!f(a)G(a)+\varepsilon_{1}(h)\;=\;\int_{a}^{b}\!f(x)g% (x)\,dx+\varepsilon_{2}(h)+G(\xi_{h})[f(b)\!-\!f(a)\!+\!\varepsilon_{3}(h)].$ (7)

Because  $f(b)\!-\!f(a)\neq 0$,  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.

Title second integral mean-value theorem SecondIntegralMeanvalueTheorem 2013-03-22 18:20:21 2013-03-22 18:20:21 pahio (2872) pahio (2872) 12 pahio (2872) Theorem msc 26A48 msc 26A42 msc 26A06