### Video Transcript

Find all π such that π of π equals the average value of π of π₯ equals π₯ minus two all squared over the closed interval negative one to five.

We begin by recalling the formula for the average value of a function over some closed interval π to π. Itβs one over π minus π times the integral of π of π₯ with respect to π₯ evaluated between π and π. We see that π of π₯ here is π₯ minus two all squared. And our interval is from negative one to five inclusive. So we say π is equal to negative one and π is equal to five.

So the average value of this function over that closed interval is one over five minus negative one times the integral evaluated between negative one and five of π₯ minus two all squared with respect to π₯. And there are a couple of ways that we can evaluate this definite integral. We could look to distribute the parentheses. Alternatively, we spot that if we let π’ be equal to π₯ minus two, then we obtain dπ’ by dπ₯ to be equal to one, which in turn means we can say that dπ’ must be equal to dπ₯.

And we can therefore use integration by substitution. We replace π₯ minus two with π’ and dπ₯ with dπ’. But we do still need to deal with the limits. So we use our definition of π’. And we see that when π₯ is equal to five, π’ is equal to five minus two, which is of course three. We also see that when π₯ is equal to negative one, π’ is negative one minus two, which is negative three. And the average value of our function is a sixth of the integral evaluated between negative three and three of π’ squared with respect to π’. And the integral of π’ squared is π’ cubed over three. And when we substitute our limits into the expression, we get a sixth of three cubed over three minus negative three cubed over three, which is simply three.

Weβre not quite finished there. Weβre trying to find the values of π such that π of π equals the average value of π of π₯ over that closed interval, in other words, when is π of π equal to three. Well, π of π is π minus two all squared. So weβre looking to find when π minus two all squared is equal to three. So weβll solve this equation for π. We take the square root of both sides of our equation, remembering to find both a positive and negative square root of three. And then we add two to both sides. And we obtain π to be equal to two plus root three and two minus root three.