Proving Uniqueness of Integral Function in [-1,1]

[takeuchi]

Let f:[-1,1]→ℝ an integrable function such that |f(x)|$$\leq2$$.
Show that there exist x in [-1,1] such that:

integral -1 to x f(t)dt-4x=0

is unique?

 

[Dick]

First show that there exists a solution to that equation, did you do that? Once you've done that consider using the Mean Value Theorem to prove uniqueness.

 

[sutupidmath]

Let f:[-1,1]→ℝ an integrable function such that |f(x)|$$\leq2$$.
Show that there exist x in [-1,1] such that:

integral -1 to x f(t)dt-4x=0

is unique?

let $$g(x)=\int_{-1}^xf(t)dt-4x$$
Now let's test the sign of this function on at the points -1 and 1 respectively.

$$g(-1)=\int_{-1}^{-1}f(t)dt-4(-1)=4>0$$

$$g(1)=\int_{-1}^1f(t)dt-4(1)\leq0$$ Since $$-2\leq f(x)\leq 2=>0=-2x|_{-1}^1=\int_{-1}^1(-2)dx\leq \int_{-1}^1f(x)dx\leq\int_{-1}^12dx=2x|_{-1}^1=4$$

Now, according to IVT there exists a number c in the interval (-1,1) such that

$$g(c)=\int_{-1}^cf(t)dt-4c=0$$

So far we know that there is a solution. Now we need to prove its uniqueness.