The numbers of possible continuous $f(x)$ defiend on $[0,1]$ for which $I_1=\int_0^1 f(x)dx = 1,~I_2=\int_0^1 xf(x)dx = a,~I_3=\int_0^1 x^2f(x)dx = a^2 $ is/are


I have tried the following: Applying ILATE (the multiplication rule for integration) - nothing useful comes up, only further complications like the primitive of the primitive of f(x). No use of the given information either. Using the rule $$ \int_a^b g(x)dx = \int_a^b g(a+b-x)dx$$ I solved all three constraints to get $$ \int_0^1 x^2f(1-x)dx = (a-1)^2 \\ \text{or} \int_0^1 x^2[f(1-x)+f(x)]dx = (a-1)^2 +a^2 \\$$ Then I did the following - if f(x) + f(1-x) is constant, solve with the constraints to find possible solutions. Basically I was looking for any solutions where the function also follows the rule that f(x) + f(1-x) is constant. Solving with the other constraints, I obtained that f(x) will only follow all four constraints if the constant [= f(x) + f(1-x)] is 2, and a is $\frac{√3\pm1}{2}$.

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    $\begingroup$ What have you tried? Please show your attempts at this problem. $\endgroup$ Aug 26 '19 at 17:43
  • $\begingroup$ I have tried using the multiplication rule for integrals to try and reach a point where I can use the given information, but have not succeeded. $\endgroup$
    – Green05
    Aug 26 '19 at 18:17
  • $\begingroup$ Can you update the question with what you have tried? $\endgroup$
    – Axion004
    Aug 26 '19 at 18:29

Apply Integration by parts on $I_2$, you will get

\begin{equation} a = x-1 \end{equation}

Also apply Integration by parts on $I_3$, you will get

\begin{equation} a^2 = x^2 - 1 \end{equation}

Above two equation satisfy only when $x=1$ and $a=0$

If you put these values in $I_2$, you will get

\begin{equation} I_2 = \int_{0}^{1} f(x) dx = 0 \end{equation} which contradict with $I_1$ so There is no such function. Ans is (D) $0$.

  • $\begingroup$ Not possible. We apply the limits after integration, so x cannot remain. $\endgroup$
    – Green05
    Aug 27 '19 at 4:09

The simplest approach is via statistics. You're counting pdfs on $[0,\,1]$ of variance $0$. A $0$-variance variable is constant. Since the Dirac Delta is disqualified for not being a function, zero functions succeed.

  • $\begingroup$ Sorry, I have no idea what you mean. Could you attach links to these topics? I have not heard of these. $\endgroup$
    – Green05
    Aug 27 '19 at 4:13
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    $\begingroup$ @Green05 I've done that. $\endgroup$
    – J.G.
    Aug 27 '19 at 6:23

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