# If $\int_{0}^{x} f = \int_{x}^{1} f$ ...

1. Jun 7, 2015

### Euklidian-Space

1. The problem statement, all variables and given/known data
If $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $\int_{0}^{x} f(x) dx = \int_{x}^{1} f(x) dx$, $\forall x \in [0,1]$, show that $f(x) = 0$ $\forall x \in [0,1]$.

2. Relevant equations

3. The attempt at a solution

I have tried doing this by assuming $\exists c \in [0,1]$ such that $f(c) > 0$. however i don't think it is bearing any fruit. I thought maybe then i could make a partition of [0,c] and [c,1] and show that their upper sums dont equal, but I dont really know where to start to show this either.

Last edited: Jun 7, 2015
2. Jun 7, 2015

### SammyS

Staff Emeritus
You should really show what variable you integrate with respect to. Do this by including a differential along with the " ∫ " (integration) symbol.

$\ \int_{0}^{x} f \$ is very ambiguous, if not downright meaningless.

You need to write something like $\ \int_{0}^{x} f(t)\,dt \$ .

3. Jun 7, 2015

### Dick

After following SammyS suggestion you might think about what happens if you take the derivative of both sides.

4. Jun 7, 2015

### Euklidian-Space

ah so use Fund THM of calc?

5. Jun 7, 2015

Right.

6. Jun 9, 2015

### hunt_mat

I would recommend the following: Let: $$F(x)=\int_{0}^{x}f(u)du$$ and consider the integral: $$\int_{0}^{1}f(u)du$$ and think about splitting the integral in some convenient way.

7. Jun 9, 2015

### WWGD

Maybe you can start by setting $x=0$ on both sides, to find the value of the integral. Then, I don't know if you have seen Weirstrass' Approximation. Thm. for continuous functions in compact intervals, so you can approximate f witin $\epsilon$ by $a_0+a_1t+...+a_{N}t^{N}$, for Real $a_i$ and try to see what happens for different choices of $x$, since it is easy to compute (and evaluate) the integral of a polynomial. Maybe a little messy (Messi?), and you need to track the approximations, but otherwise I think it works.

8. Jun 11, 2015

### haruspex

Not really. It's a pun, and not ideal, but very commonly done. The x inside the integral is a dummy variable - as always for the variable of integration - whereas the x in the bound is not, and therefore represents something else. It's exactly the same as writing the indefinite integral $\int_{0} f(x).dx$.
Search for pun at https://www.physicsforums.com/insights/conceptual-difficulties-roles-variables/

9. Jun 11, 2015

### SammyS

Staff Emeritus
What SammyS actually said was:

10. Jun 11, 2015

### haruspex

Sorry, didn't notice the OP had been edited already, copied the format from there.

11. Jun 11, 2015

### SammyS

Staff Emeritus
No problem !

I hadn't noticed his edit either.

12. Jun 12, 2015

### Fredrik

Staff Emeritus
The method that Dick suggested (take the derivative of both sides) is the fastest one. Another good approach is to add something to both sides of the equation that makes one side independent of x, and then take the derivative of both sides.

Last edited: Jun 12, 2015
13. Jun 14, 2015

### WWGD

Sorry, please ignore my reply, it is absurdly over-complicated; much better to use the answer suggested by many of using the FTC.

14. Jun 22, 2015

### Levan

The condition is true for all x∈[0,1], therefore its true for x=0
$$\int_o^o f(x) \, dx=\int_o^1 f(x) \, dx = 0$$
$$\int_o^x f(x) \, dx + \int_x^1 f(x)\, dx =\int_x^1 f(x) \, dx +\int_x^1 f(x)\, dx$$
$$\int_x^1 f(x) \, dx +\int_x^1 f(x)\, dx = \int_o^1 f(x) \, dx = 0$$
$$2\int_x^1 f(x) \, dx = 0$$
$$\int_x^1 f(x) \, dx = 0$$
$$\int_0^x f(x) \, dx = 0$$
from here you can show that f(x) is non-negative, an then from continuity you can show that (because area on the interval is 0) f(x)=0

Last edited: Jun 22, 2015