Q:

Interpreting integrals geometrically

In Apostol's Calculus, Volume $1$, section $2.19$ in the exercises, in exercise $17$ the question was to find all real values of $x$, such that
$$\int_{0}^{x} \left(t^{3} - t \right)dt = \dfrac{1}{3} \int_{\sqrt{2}}^{x} \left( t- t^{3} \right) \; dt$$
I have solved it, arrived to the last step where,
$$\dfrac{x^{4}}{4}-\dfrac{x^{2}}{2}=-\dfrac{1}{3}\left(  \dfrac{x^{4}}{4}-\dfrac{x^{2}}{2}\right)$$
and found that 
$$x=0 \quad \text{and} \quad x=\pm \sqrt{2}$$
what I don't understand is the question later on asked to draw a suitable figure and interpet the equation geometrically, do I sketch both of the functions on the integral and shade the area of integration after I plot the $x$ values as limits? And if so, what would the limits of integration be? From $-\sqrt{2}$ to $0$ and from $0$ to $\sqrt{2}$?

A:

There aren't really two functions, are there, once you take the minus sign out?
I'd sketch the graph of the integrand you see on the LHS:

and say
$x = 0$ means the LHS vanishes trivially, while the RHS vanishes since the area below the graph from $x = 0$ to $x = 1$ equals the area above the graph from $x = 1$ to $x = \sqrt{2}$.
I'll leave you to write down similar statements for the other 2 cases.

