### Video Transcript

Find the area of the region bounded by π¦ equals π₯ cubed and π¦ equals π₯.

Before we can begin to find the area of this region, we need to know what it looks like. And in order to see that, we need to produce a sketch of these two graphs. Thatβs π¦ equals π₯ cubed and π¦ equals π₯ on the same set of axes. We know what each of these graphs looks like individually. π¦ equals π₯ cubed is a simple cubic graph and π¦ equals π₯ is a straight line with a positive gradient. But we also need to know how these two graphs are positioned relative to one another.

In order to see this, we need to find the π₯-coordinates of the points where the two graphs intersect. Now, each graph has been given in the form π¦ equals something. So we can set the two expressions for π¦ equal to one another to give an equation in π₯ only. Doing so gives π₯ cubed is equal to π₯. We can then subtract π₯ from each side of the equation, giving π₯ cubed minus π₯ equals zero. And to solve this equation, we need to take out a common factor of π₯ from each term. This gives π₯ multiplied by π₯ squared minus one is equal to zero. And in fact, we can go a step further because our second factor π₯ squared minus one is a difference of two squares. So it can be factored as π₯ minus one multiplied by π₯ plus one.

Our equation has, therefore, become π₯ multiplied by π₯ minus one multiplied by π₯ plus one is equal to zero. And to solve, we need to take each factor in turn, set it equal to zero, and solve the resulting linear equation. We have first of all π₯ equals zero, then π₯ minus one equals zero, leading to π₯ equals one, and finally π₯ plus one equals zero, leading to π₯ equals negative one. And so, we find that these two graphs intersect at three places with π₯-coordinates of zero, one, and negative one.

Now, itβs just worth mentioning that it is really important that we solve this equation by first factoring π₯ from our two terms. And we donβt instead divide the equation by π₯ to give π₯ squared minus one equals zero. If we had done this, then the only solutions would be π₯ equals positive and negative one. And we would have lost the solution π₯ equals zero. So make sure you bear that in mind. We donβt divide by π₯. We factor by π₯ if we can. And therefore, we make sure that we donβt lose any solutions.

Now that we know the π₯-coordinates of the three points of intersection of these two graphs, we can sketch them on the same set of axes. We sketch π¦ equals π₯ cubed first of all. And then, we know that the graph π¦ equals π₯ needs to intersect this graph in three places: when π₯ equals negative one, when π₯ equals zero, and when π₯ equals one. We also know that π₯ cubed will be greater than π₯ for values of π₯ greater than one. So the graph of π¦ equals π₯ cubed will be above the graph of π¦ equals π₯ when π₯ is greater than one. So we add the line π¦ equals π₯ to our sketch.

And now, we can see the area of the region that weβre looking for. Itβs the area of the full region enclosed by the two graphs. So the area now shaded in green. Now, notice that this area is made up of two distinct regions. And whatβs different is that in region one, the graph of π¦ equals π₯ cubed lies above the graph of π¦ equals π₯. Whereas in region two, the graph of π¦ equals π₯ lies above the graph of π¦ equals π₯ cubed. Letβs consider region two first of all. Now, we know that in order to find the area enclosed by a line and a curve, we can use integration. And what we need to do is subtract the equation of the lower curve from the equation of the upper curve.

In this case, π¦ equals π₯ is the upper curve or upper line. And so, we have the integral of π₯ minus π₯ cubed with respect to π₯. The limits for this integral are the π₯-values where the two curves or the curve in the line intersect. So the limits will be zero and one. And we have that the area of region two is given by the definite integral from zero to one of π₯ minus π₯ cubed with respect to π₯. We can find the area of region one in a similar way. This time the limits are going to be negative one and zero. And here, the graph of π₯ cubed is above the graph of π₯. So we have the integral from negative one to zero of π₯ cubed minus π₯ with respect to π₯.

Letβs perform integral two first of all. We recall that in order to integrate powers of π₯ not equal to negative one, we increase the power by one and then divide by the new power. So we have that the integral is equal to π₯ squared over two minus π₯ to the fourth power over four evaluated between zero and one. Now, this will all be equal to zero when π₯ is equal to zero. So substituting the limits, we just have one squared over two minus one to the fourth power over four, which is a half minus a quarter which simplifies to one-quarter. So we found the area of region two.

And we can find the area of region one. By evaluating its integral this time we have π₯ to the fourth power over four minus π₯ squared over two evaluated between negative one and zero. That will all give zero when zero is substituted for π₯. So we have zero minus negative one to the fourth power over four minus negative one squared over two. Thatβs zero minus one-quarter minus a half or zero minus negative a quarter, which is equal to one-quarter. And so, we find that the area of region one is the same as the area of region two.

In fact, we could have used the rotational symmetry of the graphs of both π¦ equals π₯ and π¦ equals π₯ cubed in order to see this. And so, we could have just found the area of one of these regions and then doubled it in order to find the total area. But if not, we can just add our two separate areas together: one-quarter plus one-quarter, which is equal to one-half.

So using definite integration, we were able to find that the area of the region bounded by the curve π¦ equals π₯ cubed and the straight line π¦ equals π₯ is one-half.