# Math Genius: Prove that \$exists a, bin(0,1)\$ such that \$int_0^{a} xf(x)dx=0text{ and }int_0^bxf(x)dx=frac{b^2f(b)}{2}.\$

Question: Let $$f:[0,1]tomathbb{R}$$ be a continuous function such that $$f(0)=0$$ and $$int_0^1f(x)dx=0.$$ Prove that $$exists a, bin(0,1)$$ such that

$$int_0^{a} xf(x)dx=0text{ and }int_0^bxf(x)dx=frac{b^2f(b)}{2}.$$

My approach: Let $$g:[0,1]tomathbb{R}$$ be such that $$g(x)=xint_0^xf(t)dt-int_0^xtf(t)dt, forall xin[0,1].$$

By the first fundamental theorem of calculus we can conclude that $$g$$ is differentiable on $$[0,1]$$ and $$g'(x)=int_0^xf(t)dt, forall xin[0,1].$$

Also, observe that $$g(0)=0$$ and $$g(1)=-int_0^1tf(t)dt$$. Thus, by applying MVT to the function $$g$$ on the interval $$[0,1]$$, we can conclude that, $$exists cin(0,1)$$ such that $$g'(c)=int_0^cf(t)dt=-int_0^1tf(t)dt.$$

Observe that clearly three cases are possible, i.e, either$$int_0^cf(t)dt<0text{ or }int_0^cf(t)dt=0text{ or }int_0^cf(t)dt>0.$$

Now let $$h:[0,1]tomathbb{R}$$ be such that $$h(x)=int_0^xtf(t)dt, forall xin[0,1].$$

Please note that the part highlighted below is wrong, but still I have included it to just demonstrate my thinking process, as it might be of some help to others trying out this problem.

Observe that if $$int_0^cf(t)dt<0$$, then, $$h(1)>0$$. This also implies that $$exists$$ an open interval $$(d,e)in[0,c]$$, such that $$f(t)<0, forall tin(d,e)$$. Now select any point $$c_1in(d,e)$$. Applying MVT to the function $$h$$ on the interval $$[0,c_1]$$, we can conclude that $$exists c_2in(0,c_1)$$ such that $$h'(c_2).c_1=f(c_2).c_2.c_1=h(c_1)-h(0)=h(c_1)<0.$$

Now $$h(c_1)<0$$ and $$h(1)>0$$. Thus, by IVT we can conclude that $$exists ain(c_1,1)subseteq(0,1)$$, such that $$h(a)=int_0^af(t)dt=0.$$

A similar reasoning for the case when $$int_0^cf(t)dt>0,$$ shows that $$exists ain(0,1)$$, such that $$h(a)=int_0^af(t)dt=0.$$

Now finally if $$int_0^cf(t)dt=0$$, then we will have $$h(1)=0$$. Now if $$f$$ is identically equal to $$0$$ on $$[0,c]$$, then clearly $$tf(t)=0, forall tin[0,c]implies h(x)=0, forall xin[0,c].$$ Thus choosing any point $$xin(0,c]$$ and setting it as $$a$$, we will have $$h(a)=0$$ and we will be done in that case.

Now if $$f$$ acquires both positive and negative values on $$[0,c]$$, then we can conclude that $$exists c_1,c_2in(0,c)$$, such that $$f(c_1)>0$$ and $$f(c_2)<0$$. Also, let us assume WLOG that $$c_2>c_1$$.

I have not been able to make any significant approach other than this. Can someone help me out with this problem? Please note that a solution using integration by parts might not be possible, since $$f$$ is not a differentiable function.

The first part is answered in Olympiad calculus problem. We can build on Christian Blatter’s answer to solve the second part.

Define $$F(x) = int_0^x f(t) , dt$$ and $$phi(x) = frac 1x int_0^x F(t), dt$$, $$phi(0) = 0$$. In Christian’s answer it is demonstrated that
$$phi'(x) = frac{1}{x^2} int_0^x tf(t) , dt$$
has a zero $$a in (0, 1)$$, so that $$int_0^a tf(t) , dt = 0$$.

In our case we additionally have $$f(0) = 0$$, which implies that $$phi$$ is differentiable at $$x=0$$ with $$phi'(0) = 0$$. Applying the mean-value theorem to $$phi’$$ gives that
$$phi”(x) = -frac{2}{x^3} int_0^x tf(t) , dt + frac{f(x)}{x}$$
has a zero $$b in (0, a)$$, so that $$int_0^b tf(t) , dt = frac{b^2f(b)}{2}$$.

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