Can a Point Exist in an Integratable Function Where the Integral Equals Half?

[Dethrone]

A function is integratable on [0,1] and that $$f(x)\ge 1 $$ for $$0 \le x\le1$$. Show that there must be a point, c, on the interval [0,1] such that $$\int_{0}^{c} f(t)\,dt = \frac{1}{2}$$.

Can anyone point me to the right direction as to how to solve this? I.e does it require the IVT, the FTC, etc?

 

[Evgeny.Makarov]

This requires the IVT.

 

[Dethrone]

Just what I didn't want to hear, lol. I haven't done any proofs requiring IVT. Can you start me off?

 

[Euge]

Hi Rido12,

Since $f$ is integrable on $[0,1]$, the function

$\displaystyle F(x) = \int_ 0^x f(t)\, dt - \frac{1}{2}$

is continuous on $[0,1]$. Show that $ F (0) < 0$ and $ F (1) > 0$ (using the fact that $ f \ge 1$ on $[0,1] $). Then you can use IVT to deduce the result.

 

[Dethrone]

$\displaystyle F(x) = \int_ 0^x f(t)\, dt - \frac{1}{2}$

I know that to be true, in general:
$\displaystyle \int_{0}^{x} f(t)\,dt = F(x) + K$
because the derivative of a constant is 0, but how did you deduce the above?

 

[Euge]

I know that to be true, in general:
$\displaystyle \int_{0}^{x} f(t)\,dt = F(x) + K$
because the derivative of a constant is 0, but how did you deduce the above?

There's no need to involve the derivative. You know $ F$ is continuous on $[0,1] $, $F (0) = -\frac{1}{2} < 0$ and (since $f \ge 1$)

$\displaystyle F (1) \ge 1 - \frac{1}{2} = \frac{1}{2} > 0$.

So by the IVT, there is a point $c\in (0,1) $ such that $ F (c) = 0$. In other words,

$\displaystyle \int_0^c f (t)\, dt = \frac{1}{2} $.

 

[Dethrone]

Right! The answer was so obvious...I got up to the point right before you applied the IVT, but I thought I was oversimplifying it. Thanks!

EDIT: On second though, I think my "thanks" was a bit premature. I understand your logic, but I'm still confused as to how you came up with this conclusion:

$\displaystyle F(x) = \int_ 0^x f(t)\, dt - \frac{1}{2}$

I understand how you got $F(0)$:

$\displaystyle F(x) = \int_ 0^0 f(t)\, dt - \frac{1}{2}=0-\frac{1}{2}=\frac{-1}{2}<0$

But how did you get $\displaystyle F (1) \ge 1 - \frac{1}{2} = \frac{1}{2} > 0$?

 

[Euge]

Right! The answer was so obvious...I got up to the point right before you applied the IVT, but I thought I was oversimplifying it. Thanks!

EDIT: On second though, I think my "thanks" was a bit premature. I understand your logic, but I'm still confused as to how you came up with this conclusion:

$\displaystyle F(x) = \int_ 0^x f(t)\, dt - \frac{1}{2}$

I understand how you got $F(0)$:

$\displaystyle F(x) = \int_ 0^0 f(t)\, dt - \frac{1}{2}=0-\frac{1}{2}=\frac{-1}{2}<0$

But how did you get $\displaystyle F (1) \ge 1 - \frac{1}{2} = \frac{1}{2} > 0$?

Since $f(x) \ge 1$ for all $ x\in [0,1]$, we have

$\displaystyle \int_0^1 f(x)\, dx \ge \int_0^1 1 dx = 1$,

and therefore $ F(1) \ge 1 - \frac{1}{2} = \frac{1}{2} $.

Keep in mind this important property of integrals. If $ f $ and $ g $ are integrable on $[a, b] $ such that $f (x ) \ge g (x)$ for all $ x\in [a, b] $, then

$\displaystyle \int_a^b f (x)\, dx \ge \int_a^b g (x)\, dx$.