### Video Transcript

Find the total area bounded by the curve π¦ equals π₯ cubed in the straight line π¦ equals π₯.

Letβs begin by first drawing a sketch of the curve in the straight line. So here we have the curve π¦ equals π₯ cubed in pink and the line π¦ equals π₯ in orange. And we are asked to find the total area bounded by the curve in the line, so the total area in these two places. The curve and the line intersect at three places, π₯ equals negative one, π₯ equals zero, and π₯ equals positive one. Now, since this is a sketch, we know these values because when we have the line π¦ equals π₯, when π₯ is one, π¦ is one. And then, when we have the curve π¦ equals π₯ cubed, when π₯ is one, one cubed is one.

But to be sure, we can take these two equations and set them equal to each other and solve. This will tell us where they intersect when theyβre equal. So we can set π₯ cubed equal to π₯. Now to solve, letβs go ahead and subtract π₯ from both sides of the equation. This way, we have π₯ cubed minus π₯ equals zero. And we can solve by factoring. So we take out the greatest common factor of π₯. And when we do that, weβre left with π₯ squared minus one. Well, this is something called a difference of squares, sometimes known as D.O.T.S., difference of two squares. So when we have something such as π squared minus π squared, it factors to be π plus π times π minus π. So we take the square root of π squared and the square root of π squared. The square root of π₯ squared is π₯. And the square root of one is one. So we have π₯ times π₯ plus one times π₯ minus one.

We set each factor equal to zero and solve. So after setting them equal to zero, we have that π₯ equals zero, π₯ equals negative one, and π₯ equals positive one, just like we found in our picture. So we find the area where π₯ is less than or equal to one but greater than or equal to zero. And we also find the area where π₯ is less than or equal to zero but greater than or equal to negative one. However, since these are symmetrical, the areas in each of these little pieces will be the exact same. So if we find the area of one of them, we can either double it, which means just multiply it by two, or add it to itself.

So how are we going to find the area bounded by the curve in the straight line? We will do this by integration, from zero to one. So we take the π¦ equals π₯ line, just where we get π₯, and then subtract the π¦ equals π₯ cubed curve, which is where we get the minus π₯ cubed. So to find this integral, weβre finding the antiderivative. And the antiderivative of π₯ would be one-half π₯ squared because if we use the power rule, meaning finding the derivative, weβll bring the two down to the one-half. And two times one-half is one. And then, weβll have π₯ to the one power, making π₯.

Again, for the power rule, you bring the power down and multiply it to whatever the number is. Thatβs in front. So we have two times one-half. And then, we subtract one from the power. And two minus one is one. And then, two times one-half is one. And one times π₯ to the first power is simply π₯. So now for π₯ cubed, we will need one-fourth times π₯ to the power of four. And we evaluate at one and zero and subtract them. So as we said, we evaluate at one. And we evaluate at zero. And we will subtract them.

So letβs begin with evaluating one. One squared is one. And one times one-half is one-half. And then, one to the power of four is one. And one times one-fourth is one-fourth. Now, for zero, zero squared is zero. And zero times one-half is zero. And then again, zero to the power of four is zero times one-fourth is zero. So now, we need to simplify more. So in the first set of brackets, we have one-half minus one-fourth. We can change the one-half so has a denominator of four, meaning we have to multiply the numerator and denominator both by two, creating two-fourths. Two-fourths minus one-fourth is simply one-fourth because we take two minus one in the numerator to get one. And then, we keep our common denominator four. Zero minus zero is zero. And then, if we are saying that we subtract zero, one-fourth minus zero is simply one-fourth.

So what this means? Itβs that the area bounded by the curve π¦ equals π₯ cubed in the straight line π¦ equals π₯ is one-fourth times two, because we have one-fourth here. But donβt forget this area is also one-fourth. So as we said, we can take one-fourth times two, which we multiply the numerators together, we get two and denominators together, we get four, so two-fourths. But it reduces to one-half. So as we said, we could either take one-fourth times two or just take one-fourth plus one-fourth. And one-fourth plus one-fourth is two-fourths, because we add their numerators and keep the denominators, which reduces to one-half.

Therefore, the area bounded by the curve π¦ equals π₯ cubed in the straight line π¦ equals π₯ would be one-half.