# Find functions following this constraint

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

$$(\text{A})~~1~~~(\text{B})~~2~~(\text{C})~~\infty~~(\text{D})~~0$$

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 $$\frac{√3\pm1}{2}$$.

• What have you tried? Please show your attempts at this problem. Aug 26 '19 at 17:43
• 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. Aug 26 '19 at 18:17
• Can you update the question with what you have tried? Aug 26 '19 at 18:29

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

$$$$a = x-1$$$$

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

$$$$a^2 = x^2 - 1$$$$

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

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

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

• Not possible. We apply the limits after integration, so x cannot remain. 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.

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