### Video Transcript

Find the area of the region bounded by π¦ equals the square root of π₯ minus five and π₯ minus three π¦ equals three.

Letβs begin this question by drawing a quick sketch of these functions so that we can see what weβre doing. So the area bounded by these two equations is this area here. Well, one thing we do know that will help us is that we can find the area between a curve on the π₯-axis in the interval between π and π using integration. But in this case, we donβt want the whole area between the curve and the π₯-axis. We want the area in between the line and the curve. But if we can work out the whole area under the orange curve between the points π and π and we can work out the area under the pink line between π and π, we can then subtract one from the other to just get the area in between.

But what are these two points π and π? Well, we know itβs two points where π¦ equals the square root of π₯ minus five intersects with π₯ minus three π¦ equals three. Letβs do this by rearranging this equation for π₯ and then substituting it into this equation for π¦. Rearranging for π₯, we have that π₯ equals three π¦ add three. And substituting that into π¦ equals the square root of π₯ minus five gives us π¦ equals the square root of three π¦ add three minus five, which just simplifies to π¦ equals the square root of three π¦ minus two.

So we solve this for π¦. We should end up with two different values of π¦. And these will be the π¦-values of the points of intersection of the curve and the line. From there, weβll be able to use substitution to find the π₯-values of the points of intersection.

So letβs begin by squaring both sides of this equation in order to solve for π¦. This gives us that π¦ squared is equal to three π¦ minus two. And by rearranging, we have that π¦ squared minus three π¦ add two equals zero. And we see that we have a quadratic equation. So letβs solve this quadratic by factorizing.

We can factorize this to π¦ minus two multiplied by π¦ minus one equals zero. And that gives us our two solutions of π¦ equals two and π¦ equals one. Letβs substitute this into π₯ equals three π¦ add three to find the two values of π₯. When π¦ is equal to two, we find that π₯ is equal to nine. And when π¦ is equal to one, we find that π₯ is equal to six. So that gives us our two points of intersection of the curve and the line six and nine. So those are the two values that weβre going to need to integrate between.

Remember, we said that weβre going to need to find the area between the curve and the π₯-axis of one function and between the line and the π₯-axis of the other function. And we can then subtract these in order to find the area in between. The first function is usually the upper function, which in this case is going to be the orange curve, which is the graph of π¦ equals the square root of π₯ minus five. So the second function is going to be the function π₯ minus three π¦ equals three. But weβre going to need to rearrange this to be a function of π₯ in terms of π¦.

So rearranging, we have that three π¦ equals π₯ minus three so that π¦ equals one-third π₯ minus one. Letβs start with the integral between six and nine of the square root of π₯ minus five with respect to π₯. Because this is a root function, letβs integrate this using integration by substitution. If we choose to make the substitution π’ equals π₯ minus five, then differentiating this with respect to π₯ gives us that dπ’ by dπ₯ is equal to one.

Now, although dπ’ by dπ₯ is not a fraction, we do treat it a little bit like a fraction when weβre doing integration by substitution. So by rearrangement, we find that dπ’ is equal to dπ₯. This means that we can make a straight swap for dπ₯ with dπ’. So letβs firstly find this integral without the limits first, and then weβll include the limits afterwards.

We start by writing our integral with the substitution so that our integral is the integral of the square root of π’ dπ’. But we know that we can rewrite the square root of π’ as π’ to the power of one-half. So now we integrate π’ to the half power with respect to π’. We know that we can do this by adding one to the power to get the new power of three over two and then dividing by the new power, so dividing by three over two. And at the moment, we donβt have any limits here, so we add a constant of integration πΆ.

Now, dividing by three over two is the same as multiplying by two over three. So we can rewrite this as two over three π’ to the three over two power plus πΆ. But we remember that we let π’ equal π₯ minus five. So weβll replace π’ with π₯ minus five. Now, what we actually want here is the integral between six and nine of the square root of π₯ minus five with respect to π₯. So we need to evaluate the integral we found, two over three π₯ minus five to the three over two power, with the limits six and nine.

Note that when we have limits of integration, we donβt then need the constant of integration. So we substitute in π₯ equals nine and we substitute in π₯ equals six, and we subtract them. We can then simplify whatβs in the parentheses. And we then calculate this to be 14 over three. So thatβs the integral between six and nine of the square root of π₯ minus five with respect to π₯.

But we also need to find the integral between six and nine of one-third π₯ minus one with respect to π₯. We find this integral in the usual way, term by term, by adding one to the power and dividing by the new power. So one-third π₯ integrates to one-third π₯ squared over two. We also know that one integrates to π₯ with respect to π₯. So we need to evaluate one-third π₯ squared over two minus π₯ between six and nine. But letβs first of all simplify one-third π₯ squared over two.

Well, one-third divided by two just gives us one-sixth. And we can then substitute in π₯ is equal to nine and π₯ is equal to six and subtract them. This gives us 81 over six minus nine minus 36 over six minus six. And we find that this gives us nine over two. So we found the area under the orange curve to be 14 over three and the area under the pink line to be nine over two.

Remember, we said that to find the area in between the curve and the line, we need to subtract these. So the area in between is equal to 14 over three minus nine over two. And this just gives us one over six.