# Prove that there exists $c\in[0,1]$ such that $\int_0^cf(t)dt=f(c)^3.$

Question: Let $$f:[0,1]\to\mathbb{R}$$ be a continuous function with $$\int_0^1f(t)dt=0$$. Prove that there exists $$c\in[0,1]$$ such that $$\int_0^cf(t)dt=f(c)^3.$$

Solution: Let $$g:[0,1]\to\mathbb{R}$$ be such that$$g(x)=\int_0^xf(t)dt-f(x)^3, \forall x\in[0,1].$$

Now since $$f$$ is continuous $$\forall x\in[0,1]$$, thus, by the first fundamental theorem of calculus, we can conclude that $$g$$ is continuous $$\forall x\in[0,1]$$.

Thereafter, observe that $$g(x)=0$$ for some $$x\in[0,1]\iff \int_0^xf(t)dt=f(x)^3$$ for some $$x\in[0,1]$$. Hence, to prove the statement of the problem it is sufficient to show that $$g(c)=0$$ for some $$c\in[0,1]$$.

Now $$g(0)=-f(0)^3$$ and $$g(1)=-f(1)^3$$.

Observe that if $$f(0)$$ and $$f(1)$$ are of different signs, then $$g(0)$$ and $$g(1)$$ are also of different signs, in which case, by IVT we can conclude that $$\exists c\in(0,1)\subset[0,1],$$ such that $$g(c)=0$$. Hence, we are done in this case.

Again, if $$f(0)=0$$ or $$f(1)=0$$, then at least one of $$g(0)$$ and $$g(1)=0$$, in which case we are done.

Now, we are left with the case that both $$f(0)$$ and $$f(1)$$ are of the same sign. Thus, let us assume WLOG that $$f(0)>0$$ and $$f(1)>0$$. Hence, $$g(0)<0$$ and $$g(1)<0$$. Now since $$\int_0^1f(t)dt=0$$ and $$f(0),f(1)>0$$, implies that $$\exists$$ at least two points $$a,b\in(0,1)$$, such that $$b>a$$ satisfying $$f(a)=f(b)=0$$. Thus, we can conclude that $$\exists c_1\in(0,1),$$ such that $$f(x)>0, \forall x\in[0,c_1)$$ and $$f(c_1)=0$$. Hence, we have $$g(c_1)=\int_0^{c_1}f(t)dt-f(c_1)^3=\int_0^{c_1}f(t)dt>0.$$ Thus, we have $$g(c_1)>0$$ and $$g(1)<0$$, which implies that, by IVT we can conclude that $$\exists c\in(c_1,1)\subset[0,1]$$, such that $$g(c)=0$$. Hence, we are done in this case too.

Hence, we are done with all the cases and in each case we have shown that $$\exists c\in[0,1]$$ such that $$g(c)=0$$. Thus, we are done.

Is this solution correct and rigorous enough? If yes, is there any alternative solution?

• Looks fine and clearly explained without omitting any details. +1 for your efforts. May 14 '20 at 9:32

Appears correct and mostly rigorous to me, also clear and not too long. A proof by IVT is a valid idea. Two points:

1. Are you sure that you're applying the second FTOC, not the first FTOC?
2. You say that there exists $$0 such that $$f(x)>0$$ for all $$0\leq x < c_1$$ and $$f(c_1)=0$$. How do you know the $$f(x) > 0$$ for all $$0\leq x < c_1$$ part holds?

For part 2., you need to essentially show that $$f$$ has a smallest positive zero (assuming, for instance, that $$f(0)>0$$). Can you do it?

As a final note, the proof you gave allows one to slightly generalise the result. Namely, you can use any continuous $$h: [0, 1] \to \mathbb{R}$$ that preserves sign at $$f(0), f(1)$$ with $$h(0) = 0$$ instead of the cube function, i.e. instead of $$f(c)^3$$ you could put $$h(f(c))$$ without any trouble. Here is a slightly altered proof which, similarly to your proof, works for this generalised case (with details to be filled in by reader). Sign-preserving of $$h$$ is irrelevant at $$f(1)$$ for this proof, hence may be omitted.

Proof. For concreteness, let $$f(0) > 0$$. Let $$x_1$$ be the smallest positive zero of $$q(x) := \int\limits_{0}^{x}f(t)\,\mathrm{d}t$$. We may assume $$f(x_1) < 0$$. Then by IVT at some point $$z$$ in $$(0, x_1)$$ it is the case that $$f(z) = 0$$ and $$q(z) > 0$$. Therefore, $$g := q - h(f)$$ will have changed sign in $$(0, z)$$, completing the proof.

• Fantastic, +1, I always knew there was a generalization and an easier proof!! May 14 '20 at 16:44