# Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $\int_{x}^{1}f(t)dt\geq(1-x)^{2}$. Prove that $f(1)=0$

Is this a proof? The problem states: Let $$f:\mathbb{R}\rightarrow\mathbb{R}$$ be a continuous function such that $$\int_{x}^{1}f(t)dt\geq(1-x)^{2}$$, for any $$x\in\mathbb{R}$$. Prove that $$f(1)=0$$.

I denoted by $$F(x)=\int_{x}^{1}f(t)dt$$ and I noticed that $$F(1)=0$$ and $$F(x)\geq0=F(1)$$ and now, $$f(1)=\lim_{x\rightarrow1,x>1}\frac{F(x)-F(1)}{x-1}\geq0$$ and $$f(1)=\lim_{x\rightarrow1,x<1}\frac{F(x)-F(1)}{x-1}\leq0$$, so $$f(1)=0$$. Is this correct?

• Looks fine to me. Mar 6, 2019 at 5:27
• Well, $F(x)\geqF(1)$ and from there result the inequalities.
– user318394
Mar 6, 2019 at 7:04

This is quite elementary

for $$x\leq1$$ we have

the area under $$f(x)$$ is $$\geq0$$ since $$(1-x)^2\geq0$$

$$\implies f(t)\geq0$$ for $$x<1$$

for $$x\geq1$$ we switch our limits giving us a negative negative sign our equation becomes as follows

$$-\int_{1}^{x}f(t)dt\geq(1-x)^{2}$$,

multiplying with a negative sign we get

$$\int_{1}^{x}f(t)dt\leq-(1-x)^{2}$$,

the area under $$f(x)$$ is $$\leq0$$ since $$-(1-x)^2\leq0$$

$$\implies f(t)\leq0$$ for $$x>1$$

using intermediate value theorem since $$f(x)$$ is greter than 0 for $$x<1$$ and less than 0 for $$x>1$$

$$\implies f(1)=0$$