[GRE 1268, #14a] Suppose that \(f\) is a continuous function, and \(a,b\) are real numbers such that \[\int_{a}^x f(t)dt = \int_a ^b f(t)dt+24+3x^3 \hspace{2ex} \mbox{ for all } x.\] Find \(b\).

Solution At first sight, this problem looks more difficult than it is. The unknown function \(f\) is a big distractor. But the key to the problem is in the words '\(\mbox{for all } x\)' in the given equation. That means that we can choose any \(x\) we want for that equation. So look for a choice that will make the unknown function \(f\) disappear. We notice that the two integrals are on different sides of the equation and they are the same except that the upper boundary is \(x\) for one and \(b\) for the other. So we use \(x=b\). Then the two integrals are the same and they cancel. This means that \[0=24+3b^3.\] Solving for \(b\) we find \(b= -2\).
There is another, longer way to solve this problem: you could take the derivative of both sides of the given equation and use the Fundamental Theorem of Calculus (Part I): \[\frac{d}{dx}\int_a^x f(t)dt = f(x).\] Using the fact that the definite integral on the right is a number (and so its derivative is zero), this will allow you to find that \(f(x) = 9x^2\). You can then substitute this back in the given equation, do both integrals, and after some work and simplification you will find the same equation: \(0=24+3b^3\). But clearly the first way to do it is better.