Number of Functions to Satisfy $\alpha$ Int. Conditions

[juantheron]

Nymber of all positive continuous function $f(x)$ in $\left[0,1\right]$ which satisfy $\displaystyle \int^{1}_{0}f(x)dx=1$ and $\displaystyle \int^{1}_{0}xf(x)dx=\alpha$ and $\displaystyle \int^{1}_{0}x^2f(x)dx=\alpha^2$Where $\alpha$ is a given real numbers.My trial:: Using Addition and Subtraction, I am getting $\displaystyle \int^{1}_{0}(x-1)^2f(x)dx=(\alpha-1)^2$ now how can I solve it, Help required, Thanks

 

[jvicens]

Thank you for your post. I am happy to help you with your question. First, let's rewrite the given conditions in a more compact form:

$$\int_0^1 f(x) dx = 1$$
$$\int_0^1 xf(x) dx = \alpha$$
$$\int_0^1 x^2f(x) dx = \alpha^2$$

Now, let's look at the integral $\int_0^1 (x-1)^2f(x)dx$. This can be expanded to $\int_0^1 (x^2 - 2x + 1)f(x)dx$. Using the given conditions, we can rewrite this as:

$$\int_0^1 (x^2 - 2x + 1)f(x)dx = \int_0^1 x^2f(x)dx - 2\int_0^1 xf(x)dx + \int_0^1 f(x)dx$$

Substituting in the given values for the integrals, we get:

$$\int_0^1 (x^2 - 2x + 1)f(x)dx = \alpha^2 - 2\alpha + 1$$

Now, since we also know that $\int_0^1 (x-1)^2f(x)dx = (\alpha-1)^2$, we can equate the two expressions and solve for $f(x)$:

$$(\alpha-1)^2 = \alpha^2 - 2\alpha + 1$$
$$\alpha^2 - 2\alpha + 1 = \alpha^2 - 2\alpha + 1$$

Therefore, we can see that $f(x)$ can take on any positive value as long as it satisfies the given conditions. In other words, there are an infinite number of positive continuous functions $f(x)$ that satisfy the given conditions.

I hope this helps answer your question. Let me know if you have any further questions or need clarification.