Find all x such that the integral equation holds

Find all values of $x \in \mathbb{R}$ so that

$\int_0^x (t^3 - t) \, dt = \frac{1}{3} \int_{\sqrt{2}}^x (t - t^3) \, dt.$

First, we evaluate the integral on the left,

$\int_0^x (t^3 - t) \, dt &= \frac{x^4}{4} - \frac{x^2}{2}.$

Then, we evaluate the integral on the right,

$\frac{1}{3} \int_{\sqrt{2}}^x (t - t^3) \, dt = -\frac{1}{3} \left( \frac{x^4}{4} - \frac{x^2}{2} \right)$

So, we then set these equal,

$\int_0^x (t^3 - t) \, dt = \int_{\sqrt{2}}^x (t - t^3) \,dt \quad \implies \quad \frac{x^4}{4} - \frac{x^2}{2} &= -\frac{1}{3} \left( \frac{x^4}{4} - \frac{x^2}{2} \right)$

We can see by substituting that $x = 0$ is one solution. If $x \neq 0$ then we may divide by $x^2$ and obtain,

$\begin{align*} \frac{x^2}{4} - \frac{1}{2} = -\frac{1}{3} \left( \frac{x^2}{4} - \frac{1}{2} \right) && \implies && \frac{x^2}{3} &= \frac{2}{3} \\ && \implies && x^2 &= 2 \\ && \implies && x = \pm \sqrt{2}. \end{align*}$

Thus, all solutions are given by $x = 0, \sqrt{2}, -\sqrt{2}$ .

Stumbling Robot

Find all x such that the integral equation holds

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