### Video Transcript

Find the area of the region enclosed by the curves π¦ equals five π₯ and π¦ equals two π₯ minus five all squared.

With questions of this type, you could use graph plotting software to observe the bounded region. However, letβs instead work on finding the important points on our curves algebraically, starting with π¦ equals two π₯ minus five all squared. Looking at our equation, we can see that the leading term will be some coefficient times π₯ squared. And we can therefore say that our graph shape will be that of a parabola.

To better understand this curve, let us now work on finding the π¦-axis intercept. For both of our equations, we have been given π¦ in terms of π₯. We know that when our curve crosses the π¦-axis, π₯ will be equal to zero. We can input this π₯-value into our equation in order to find the π¦-value. Solving this equation, we find that π¦ is equal to 25 when π₯ is equal to zero. We can therefore say that the π¦-axis intercept is at the point zero, 25.

Now let us continue by finding our π₯-axis intercepts. We know that when our curve crosses the π₯-axis, π¦ will be equal to zero. Since we have been given π¦ in terms of π₯, we can add this relationship to our equation. Inspecting this equation, we can see that we have a set of parentheses, squared, is equal to zero. The only way that this can be true is if the set of parentheses itself is also equal to zero.

We can therefore proceed by saying two π₯ minus five is equal to zero. We can now solve the equation by adding five to each side and dividing by two. We have now found that when π¦ is equal to zero, π₯ is equal to five over two. We can therefore say that we only have one π₯-axis intercept at the point five over two, zero.

Looking at our curve, we can see that there is only one horizontal line that will only cut our curve once. All other horizontal lines will lead to two intersections. Since we are working with a parabola, we can also conclude that this point, the π₯-axis intercept, is also the turning point of our curve. We are now in a better position to sketch our graph using the points that we have found.

Let us now turn our attention to the other equation, π¦ equals five π₯. We can immediately see that this equation describes a straight line with positive gradient passing through the origin. Here, we have drawn this line on our graph. Now that we have completed our graph sketch, we can shade in the region bounded by the curve and the line. Let us consider how to find the area of this region. When trying to find the area under a curve, one of the tools that we can use are definite integrals.

As a quick check, we can look at our bounded region and see that the enclosed area does not pass below the π₯-axis. From this, we can conclude that at no point will our definite integral evaluate to a negative area and we can safely proceed without splitting our region into parts. In order to proceed with this method, we will need to find the limits for our integration. Looking back at our sketch, we can see that the important points are where the line and the curve intersect.

We can use the π₯-axis coordinate of these intersection points for our limits π and π of our definite integral. In order to visualize our method more clearly, let us sketch out our line and our curve separately, both having the intersection points marked. Using π and π as the π₯-axis coordinates of these intersection points, we can see the area that would be bound by each of our curve and the π₯-axis between these limits.

We have labelled our first area π΄ one and our second area π΄ two. By inspection, we can see that subtracting π΄ two from π΄ one will give us the area of the bounded region. Let us call the area of the bounded region π΄ and keep this formula for later. Our method is then to find π΄ one or the area enclosed by the line π¦ equals five π₯ and the π₯-axis between the limits of π and π. We will then do the same for π΄ two and use our formula to find the area of the bounded region.

As a quick side note, you may notice that area π΄ one is a trapezoid and could be found using geometric methods. For this example, weβre going to choose to use integration to find the area. However, you can use other techniques if you like. Now that we understand our goal, let us work on finding our limits, small π and small π. We can do this by finding the intersection points of our line and our curve. We know that at these points the π¦ and π₯ values of the two equations will be the same. And we can therefore equate the two expressions of π₯.

Now that we have made an equation for π₯, let us work on solving, first by multiplying out the parentheses on the left-hand side. Now let us also collect our π₯ terms on the left-hand side. For our next step, we will work on factorizing this equation. At this stage, we could use the quadratic formula. However, itβs also possible to factorize directly. One clue we may take is that 25 only has factors of one, five, and itself.

We find that we are unable to successfully use one and 25 and we therefore work with the fives. With some trial and error, we can find that our equation successfully factorizes to four π₯ minus five and π₯ minus five. Now we have a familiar situation in which two brackets multiplied together equals zero. We know that in order for this situation to be true, one of our factors must also be equal to zero. We can therefore say that four π₯ minus five equals zero or π₯ minus five equals zero.

Solving these two cases, we get that π₯ is equal to five over four or π₯ is equal to five. Since we do not need the π¦ values for this question, these points can be marked on our graph now. Finally, we are ready to look at our integration starting with area π΄ one. This is the area bounded by the line π¦ equals five π₯ and the π₯-axis between our newly found limits of five over four and five. Here, we see the definite integral of five π₯ with respect to π₯ between our limits.

Using the rules of integration, this evaluates to five over two π₯ squared again between the limits. As our next step, we have taken out a factor of five over two and input our limits into the equation. Let us now work through our calculations, first by squaring our terms and simplifying. Finally, we find that π΄ one is equal to 1875 over 32. For now, let us put this value aside and work on π΄ two.

Here, we have written the definite integral for π΄ two, which is the area bound by the curve π¦ equals two π₯ minus five all squared and the π₯-axis between the limits we have found, which are five over four and five. To help our calculations, let us use the expanded form of these parentheses that we found earlier. Here, we have integrated our function raising the power of each π₯ term by one and dividing by the new power.

Let us quickly simplify 20 over two as 10. As our next step, weβre going to express 10 and 25 in terms of five. The reasons for doing this will become apparent in a moment. We are now ready to input our limits into the equation. Here, we see the full equation with our limits marked in a different color to help keep track of the numbers. With a little bit of inspection, we should be able to see that all of our terms contain a factor of five cubed.

Now it may be a little bit difficult to spot this at first, but upon taking out our factor of five cubed, we get an equation which is slightly easier to work with. We can continue to simplify our equation. In doing so, our answer eventually reduces down and we find that our area π΄ two is equal to 375 over 16. We are now ready to find the total bounded area using the formula that we defined earlier.

Taking π΄ one, the area bound by the line π¦ equals five π₯, and subtracting π΄ two, the area bound by the line π¦ equals two π₯ minus five all squared, we find the total area of our bounded region is 1125 over 32 units squared or area units.