What is the smallest possible value of the integral ∫(x-a)^2 dx between 0 and 1 as a varies?

This is a minimisation problem, but it's slightly tricky to see what it is we're minimising. Seeing as it's a that's varying, we are going to have to differentiate with respect to a at some point. We're going to need to find the value of the integral depending on a - this is what we're trying to get the smallest value of. As a first step, we can expand the bracket and make it easier to integrate. This gives us the expression (x² -xa + a²) inside the integral. Getting the anti-derivatives of this, we need to evaluate [1/3 x³ -x²a +xa] on the interval 0 to 1. Since all of the terms have an x in them, they will be 0 when x is 0, so we just have to substitute x =1. This will give us our value as a function of a, what we were trying to find! We can write f(a) = 1/3 - a - a². Now to find the minimum, we differentiate with respect to a and set to 0. This gives 2a - 1 = 0, and we find a = 1/2. Our value is given by putting 1/2 into our function f(1/2)= 1/3 -(1/2) + (1/2)² = 1/12, our answer! To check that is actually a minimum, we can see that the second derivative of f(a) is 2, which is always positive.