Can We Prove x1 Equals x2 When the Integral of a Positive Function Equals Zero?

[Bipolarity]

Consider the function $F(x)$ where $F(x) > 0$ for all $x$.

If we know that $$\int^{x_{2}}_{x_{1}}F(x)dx = 0$$ can we prove that $x_{1}=x_{2}$ ?

I can visually imagine that they are equal since the function is always positive, its integral must be monotically increasing, but I can't imagine how I would prove this.

I made the problem myself while studying probability so I'm not sure a solution exists. If a solution does not exist I'd like to see a counterexample.

I would imagine that the solution employs the MVDT and FTC, but as I mentioned before, I'm not good at actually writing the statements for proofs so I need some help here.

BiP

 

[Char. Limit]

Let G'(x)=F(x), i.e. let G(x) represent a primitive of F(x). Then the integral in your post is equal to G(x₂) - G(x₁). After this, you can use monotonicity of the derivative F(x) to prove that G(x₂) - G(x₁) = 0 implies x₂ - x₁ = 0, and your statement follows.

 

[lavinia]

Consider the function $F(x)$ where $F(x) > 0$ for all $x$.

If we know that $$\int^{x_{2}}_{x_{1}}F(x)dx = 0$$ can we prove that $x_{1}=x_{2}$ ?

I would imagine that the solution employs the MVDT and FTC, but as I mentioned before, I'm not good at actually writing the statements for proofs so I need some help here.

BiP

If the endpoints are not equal then there is a strictly positive lower Riemann sum under the curve. The integral is bounded below by this sum.