# Find function $f(x)$ that is continuous on $[0,2]$ satisfies $f(2) = 3$; $\int_0^2 [f'(x)]^2 dx = 4$ and $\int_0^2 x^2f(x) dx = \frac{1}{3}$,

How to find function $$f(x)$$ that has continuous derivative on $$[0,2]$$ satisfies the following conditions:

1. $$f(2) = 3$$
2. $$\displaystyle \int_0^2 [f'(x)]^2 dx = 4$$
3. $$\displaystyle \int_0^2 x^2f(x) dx = \frac{1}{3}$$

My attempt: By using integration by parts, I found that $$\displaystyle \int_0^2 x^3f'(x) dx = 23$$ and I tried to find constant $$\alpha$$ such that $$\displaystyle \int_0^2 [f'(x) + \alpha x^3]^2 dx = 0$$ so that I can have $$f'(x) = -\alpha x^3$$. However, the result was strange, I obtained two different "ugly" values and failed to confirm whether the solution was right. I then searched for solution online but did not come across anything helpful.

I would love to know is there another way to solve this problem. I'm grateful if anyone could help. Thanks in advance.

• There are three condtioins. Did you try \$f(x)=a+bx+cx^{2}? Feb 19, 2020 at 7:36
• Yes, I have just tried it but it still gave me weird solutions
– KM02
Feb 19, 2020 at 7:57

We have, by Cauchy - Schwarz:

\begin{align} 23 & =\int_0^2 x^3f'(x) \, dx \le \left(\int_0^2 x^6 dx\right)^{1/2} \cdot \left(\int_0^2 (f'(x))^2 \, dx \right)^{1/2} \\ & =2 \left(\int_0^2 x^6 \, dx\right)^{1/2} \le 2 \left(\int_0^2 2^6 \, dx \right)^{1/2} =\sqrt{2} \cdot 16 < 23. \end{align}

A contradiction ! Consequence ?

• Oh wow, thanks, I did not notice that. So the problem was wrong the whole time?
– KM02
Feb 19, 2020 at 9:30

$$\int_0^2 x^3f'(x)\, dx= \int_0^2 x^3 \,df(x)=x^3 f(x)|^2_0-3\int_0^2 x^2f(x) \,dx=8\cdot f(2)-3\cdot\frac{1}{3}=24-1=23.$$

• Yes, I have already mentioned it above
– KM02
Feb 19, 2020 at 7:58