,If $\int_{0}^{1}f(x)dx$=$\int_{0}^{1}g(x)dx$,then show that$ f= g $.

Question

If f and g are continuous functions on $[0, 1]$ satisfying$ f(x) ≥ g(x)$for every$ 0 ≤ x ≤ 1$,If $\int_{0}^{1}f(x)dx$=$\int_{0}^{1}g(x)dx$,then show that$ f= g $.

I don't how to prove this as i can prove if itake$ f(x)=g(x)= 0$ i think this will not be right way to direct proved

Pliz help me..

Answer 1

This problem is equivilent to: $\displaystyle\int_0^1 f(x) dx = 0 ,f(x) \ge 0$ then $f(x)=0$.

Prove by contradiction, if there is some $x \in (0,1),f(x)=t>0$,there is a neibourhood $(x-\epsilon,x+\epsilon)$ of $x$ such that $f(x) \ge \frac{t}2$(since $f(x)$ is continuous),so

$ \displaystyle \int_0^1 f(x) dx \ge \int_{x-\epsilon/2}^{x+\epsilon/2} f(x) dx \ge 2 \epsilon \dfrac{t}2 > 0$

contradiction!

Answer 2

Hint:

Write $h(x)=f(x)-g(x)$, then $h(x)$ is a positive continuous function and it satisfy that $\int_{0}^{1}h(x)dx=0$.

If (by contradiction) $h(x_0) = c >0$ what can you tell about the values of $h$ in an interval close to $x_0?$