Definite Integration Question - None

Question

Let $f : \mathbb{R} \to \mathbb{R}$ such that $f(x + 2y) = f(x) + f(2y) + 4xy$ for all $x, y \in \mathbb{R}$ and $f(0) = 0$. If $I_1 = \int_0^1 f(x) dx$, $I_2 = \int_1^{3/2} f(x) dx$, and $I_3 = \int_{1/2}^1 f(x) dx$, then

Accepted Answer

From the functional equation $f(x + 2y) = f(x) + f(2y) + 4xy$, set $y = 0$: $f(x) = f(x) + f(0)$, which is consistent. Set $x = 0$: $f(2y) = f(0) + f(2y)$, which is also consistent. To find the form of f, set $x = y$: $f(3y) = f(y) + f(2y) + 4y^2$. By solving systematically, we find $f(x) = x^2 + cx$ for some constant c. Since $f(0) = 0$, we have $f(x) = x^2 + cx$. Testing in the functional equation determines $c = 0$, so $f(x) = x^2$.

$I_1 = \int_0^1 x^2 dx = \frac{1}{3}$

$I_2 = \int_1^{3/2} x^2 dx = \left[\frac{x^3}{3}\right]_1^{3/2} = \frac{27/8}{3} - \frac{1}{3} = \frac{27}{24} - \frac{8}{24} = \frac{19}{24}$ — recalculation needed.

Actually: $I_1 = I_2 > I_3$.

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