Question Video: Finding the Area of a Region Bounded by Quadratic and Absolute Value Functions Mathematics • Higher Education

Find the area of the region bounded by 𝑦 = 2 βˆ’ |π‘₯| and 𝑦 = π‘₯⁴.

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Video Transcript

Find the area of the region bounded by 𝑦 is equal to two minus the absolute value of π‘₯ and 𝑦 is equal to π‘₯ to the fourth power.

In this question, we’re asked to find the area of a region bounded by the graphs of two given functions. 𝑦 is equal to two minus the absolute value of π‘₯ and 𝑦 is equal to π‘₯ to the fourth power. And whenever we’re asked to find the area of a region bounded by the graphs of two given functions, it’s a good idea to sketch the information we’re given. So let’s start by sketching the graph of 𝑦 is equal to two minus the absolute value of π‘₯.

The easiest way to sketch this graph is to notice that it’s a series of transformations on the absolute value graph 𝑦 is equal to the absolute value of π‘₯. First, 𝑦 is equal to negative the absolute value of π‘₯ is a reflection in the π‘₯-axis. We can then transform this graph into 𝑦 is equal to two minus the absolute value of π‘₯ by translating the graph two units vertically. This then gives us a graph which looks somewhat like the following. And although it’s not necessary, we could find the 𝑦-intercept by substituting π‘₯ is equal to zero into the function. Or we could find the values of the π‘₯-intercepts by solving the function equal to zero. However, as we’ll see, it’s not necessary for this question.

Now we need to sketch on the same coordinate axis the graph of 𝑦 is equal to π‘₯ to the fourth power. In this case, 𝑦 is equal to π‘₯ to the fourth power has a very similar shape to the curve 𝑦 is equal to π‘₯ squared. We can now sketch on our diagram the region we’re asked to find the area of. And we can see that this region is bounded above by 𝑦 is equal to two minus the absolute value of π‘₯ and below by 𝑦 is equal to π‘₯ to the fourth power. We can then recall one of the properties of integration to allow us to evaluate the area of this region.

We can recall if we have two functions 𝑓 of π‘₯ and 𝑔 of π‘₯, where 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ on a closed interval from π‘Ž to 𝑏, then the area between the curves 𝑦 is equal to 𝑓 of π‘₯ and 𝑦 is equal to 𝑔 of π‘₯ and the vertical lines π‘₯ is equal to π‘Ž and π‘₯ is equal to 𝑏 is given by the definite integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯ with respect to π‘₯. And it’s worth pointing out this is only valid provided 𝑓 and 𝑔 are integrable on the closed interval from π‘Ž to 𝑏. And we can show that all of this is true in this case. First, our value of π‘Ž and our value of 𝑏 will be the π‘₯-coordinates of the two points of intersections between the two curves. Then, on this interval, we know that two minus the absolute value of π‘₯ is greater than or equal to π‘₯ to the fourth power.

Finally, we know that both of these functions are integrable on this interval because they’re both continuous for all real values of π‘₯. Therefore, we can use this to find the given area. We just need to find the values of π‘Ž and 𝑏. To find the π‘₯-coordinates of the points of intersections of the graph of the two functions, we’ll just set the two functions equal to each other and solve for π‘₯. We need to solve two minus the absolute value of π‘₯ is equal to π‘₯ to the fourth power. We’ll do this by first rearranging the equation to make the absolute value of π‘₯ the subject. We get two minus π‘₯ to the fourth power is equal to the absolute value of π‘₯.

And now we get two possibilities depending on whether our value of π‘₯ is positive or negative. If π‘₯ is positive, we get two minus π‘₯ to the fourth power is equal to π‘₯. And if π‘₯ is negative, we get two minus π‘₯ to the fourth power is equal to negative π‘₯. And both of these equations are degree four polynomial equations. So they’re difficult to solve by using standard methods. A good idea is to try substituting in small integer values of π‘₯ to see if we can find a solution. Let’s start by trying π‘₯ is equal to one. Remember, when π‘₯ is positive, we need to use the left equation. Substituting π‘₯ is equal to one, we get two minus one to the fourth power is equal to one. Since this equation holds true, π‘₯ is equal to one is a solution to this equation. And in particular, this means it’s an π‘₯-coordinate of the points of intersection between the two curves.

Since our value of 𝑏 is positive and π‘Ž is negative, we’ve shown 𝑏 is equal to one. And now we can determine π‘Ž in two different ways. First, 𝑦 is equal to π‘₯ to the fourth power and 𝑦 is equal to two minus the absolute value of π‘₯ are symmetric about the 𝑦-axis. So the points of intersection will be reflections in the 𝑦-axis. The intersection will be at π‘₯ is equal to negative one. But we can also confirm this by substituting π‘₯ is equal to negative one into the second equation. Substituting negative one into the left-hand side of our equation, we get two minus negative one to the fourth power. Evaluating this, we get that it’s equal to one, which we can see is equal to the right-hand side of our equation when we substitute π‘₯ is equal to negative one.

Therefore, we’ve shown the area of the shaded region is the definite integral from negative one to one of two minus the absolute value of π‘₯ minus π‘₯ to the fourth power with respect to π‘₯. And now there’s a few different ways we could evaluate this integral. For example, the integrand can be written as a piecewise function. However, in this case, we’re going to split our integral over the functions 𝑓 of π‘₯ and 𝑔 of π‘₯ separately. First, by using our properties of integrals, we can rewrite this as the integral from negative one to one of two minus the absolute value of π‘₯ with respect to π‘₯ minus the integral from negative one to one of π‘₯ to the fourth power with respect to π‘₯.

Now each of these integrals represents the signed area under the graph of the function between negative one and one, although in our diagram we can see the graphs of both of these functions are above the π‘₯-axis on this interval. We can use this diagram to determine the area under 𝑦 is equal to two minus the absolute value of π‘₯ between negative one and one. And to help us do this, let’s resketch this graph. We want to determine the area of this graph between π‘₯ is equal to negative one and one. And we can do this by splitting the region into two, into a rectangle and a triangle. We can then find the area of the rectangle. Its width is two and its height is one, so its area is two.

We then need to find the area of the triangle. And one way of doing this is to note the 𝑦-intercept of the graph is at two. So the height of the triangle is one, and its base has length two. Therefore, the area of this triangle is a half its base times its perpendicular height, a half times two times one. And adding onto this value the area of the rectangle, we’ve evaluated the definite integral from negative one to one of two minus the absolute value of π‘₯ with respect to π‘₯.

However, we can’t do the same thing for the second integral. So instead, we’re going to need to use the power rule for integration. And we recall this tells us we can evaluate this integral by adding one to the exponent of π‘₯ and then dividing by this new exponent. We get π‘₯ to the fifth power over five evaluated at the limits of integration, π‘₯ is equal to negative one and π‘₯ is equal to one.

We can now start evaluating. First, a half times two times one is equal to one and two plus one is equal to three. Next, we need to evaluate our antiderivative of π‘₯ to the fifth power over five at the limits of integration. We get one to the fifth power over five minus negative one to the fifth power over five. Therefore, the area of the shaded region is three minus one to the fifth power over five minus negative one to the fifth power over five. And now we just need to evaluate this expression. One to the fifth power minus negative one to the fifth power is two to the fifth power. So we get three minus two to the fifth power, which is 13 over five.

And we could leave our answer like this. However, we’re going to write this as a mixed fraction, which is two and three-fifths. And since this represents an area, we could say that this is square units. However, it’s not necessary. Therefore, we were able to show the area of the region bounded by 𝑦 is equal to two minus the absolute value of π‘₯ and 𝑦 is equal to π‘₯ to the fourth power is two and three-fifths.

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