0
$\begingroup$

My vector calculus textbook states the following:

$$\int_a^b h(t) g(t) \ dt = h(c) \int_a^b g(t) \ dt,$$

provided $h$ and $g$ are continuous and $g \ge 0$ on $[a, b]$; here $c$ is some number between $a$ and $b$.$^3$

$^3$ Proof If $g = 0$, the result is clear, so we can suppose $g \not= 0$; thus, we can assume $\int_a^b g(t) \ dt > 0$. Let $M$ and $m$ be the maximum and minimum values of $h$, achieved at $t_M$ and $t_m$, respectively. Because $g(t) \ge 0,$

$$m \int_a^b g(t) \ dt \le \int_a^b h(t) g(t) \ dt \le M \int_a^b g(t) \ dt$$

Thus, $\dfrac{\left( \int_a^b h(t) g(t) \ dt \right)}{\left( \int_a^b g(t) \ dt \right)}$ lies between $m = h(t_m)$ and $M = h(t_M)$ and therefore, by the intermediate value theorem, equal $h(c)$ for some intermediate $c$.

Can someone please explain how $\dfrac{\left( \int_a^b h(t) g(t) \ dt \right)}{\left( \int_a^b g(t) \ dt \right)}$ lies between $m = h(t_m)$ and $M = h(t_M)$? This was simply asserted with no justification.

Thank you.

$\endgroup$
1
$\begingroup$

If $M(h(t))=\sup_{t\in[a,b]}h(t)$ and $m(h(t))=\inf_{t\in[a,b]}h(t)$,

$m(h(t))\le h(t)\le M(h(t))\forall t\in[a,b]\implies m(h(t))g(t)\le h(t)g(t)\le M(h(t))g(t)\forall t\in[a,b]$ if $g(t)\ge 0$.

$\because g(t),h(t)\in\mathfrak{R}[a,b]\implies g(t)h(t)\in\mathfrak{R}[a,b],$ where $\mathfrak{R}[a,b]$ is the set of all Riemann integrable functions defined on $[a,b]$

$\therefore \displaystyle\int_a^b m(h(t))g(t)\,dt\le \displaystyle\int_a^bh(t)g(t)\,dt\le \displaystyle\int_a^bM(h(t))g(t)\,dt$

$\implies m(h(t))\displaystyle\int_a^bg(t)\,dt\le \displaystyle\int_a^bh(t)g(t)\,dt\le M(h(t))\displaystyle\int_a^bg(t)\,dt$

$\implies m(h(t))\le \dfrac{\displaystyle\int_a^bh(t)g(t)\,dt}{\displaystyle\int_a^bg(t)\,dt}\le M(h(t))$

Similarly, if $g(t)\le 0$, $M(h(t))\le \dfrac{\displaystyle\int_a^bh(t)g(t)\,dt}{\displaystyle\int_a^bg(t)\,dt}\le m(h(t)).$

Thus $\dfrac{\displaystyle\int_a^bh(t)g(t)\,dt}{\displaystyle\int_a^bg(t)\,dt}=\mu$ where $m(h(t))\le\mu\le m(h(t))$

If in case $h(t)\in\mathfrak{C}[a,b]$ then $h$ attains all values between its supremum and infimum. So $\exists\zeta\in[a,b]:\mu=h(\zeta).$

Hence $\dfrac{\displaystyle\int_a^bh(t)g(t)\,dt}{\displaystyle\int_a^bg(t)\,dt}=h(\zeta)$ for some $\zeta\in[a,b].$

$\endgroup$
  • $\begingroup$ Thanks for the elaborate answer! :) $\endgroup$ – The Pointer Oct 17 '18 at 8:00
  • $\begingroup$ You are welcome! :) $\endgroup$ – Arjun Banerjee Oct 17 '18 at 10:22
1
$\begingroup$

It follows from the inqualities right before. Since $M$ is the maximum, $h(t)\leq M$ for all $t\in[a,b]$. Hence $$ \frac{\int_a^b h(t)g(t)\,dt}{\int_a^b g(t)\,dt} \leq \frac{M \int_a^b g(t)\,dt}{\int_a^b g(t)\,dt} = M. $$ The other inequality follows similarly from $m\leq h(t)$ for all $t\in[a,b]$.

$\endgroup$
  • $\begingroup$ Ahh, I understand now. Written slightly differently, $\frac{\int_a^b h(t)g(t)\,dt}{\int_a^b g(t)\,dt} \leq \frac{ M \int_a^b g(t)\,dt}{\int_a^b g(t)\,dt} = M = \frac{ \int_a^b M g(t)\,dt}{\int_a^b g(t)\,dt}$, which makes it clear. Thanks for the clarification, Ernie! $\endgroup$ – The Pointer Oct 15 '18 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.