91Ó°ÊÓ

When dealing with multiple curves, understanding where they intersect is important to finding the region enclosed by them. The intersection points tell us where the boundaries of the region are, which then allows us to calculate the area within those boundaries.
To find the intersection points, set the equations of the curves equal. For example, to find the intersection of the lines given by the equations \(y = x\) and \(y = 4x\), you equate the expressions: \(x = 4x\). Solve for \(x\) and you obtain \(x = 0\), leading to \(y = 0\), so one point of intersection is \((0, 0)\).
Repeat this process for the other combinations of curves: \(y = x\) vs \(y = -x + 2\), and \(y = 4x\) vs \(y = -x + 2\), giving you additional intersection points \((1, 1)\) and \(\left( \frac{2}{5}, \frac{8}{5} \right)\).

• These intersection points form the vertices of the enclosed region.
• Understanding these points is crucial for further calculations.

Once the intersection points are known, we can proceed to find the area of the region enclosed by integrating the functions. Calculus integration helps us calculate the area under a curve or between curves.
In the case of our example, we have multiple curves intersecting. The task is to compute the vertical distance between the curves over the interval determined by their intersections and then integrate these differences.
This is usually done by integrating the top function minus the bottom function of the enclosed area over the range confined by the points of intersection. For instance, if we have two functions \(f(x)\) and \(g(x)\), where \(f(x)\) is above \(g(x)\) for a certain interval, the area \(A\) is given by:
\[ A = \int _{a}^{b} (f(x) - g(x)) \, dx \]
where \(a\) and \(b\) are the \(x\)-coordinates of the points of intersection.

• The integration simplifies to calculate real-world areas using algebraic functions.
• It involves summing infinitesimal strips of width \(dx\) over a defined interval.

Coordinate geometry is fundamental for visualizing how different curves interact on a plane. In this context, it helps in sketching and understanding the spatial relationships between the given lines: \(y = x\), \(y = 4x\), and \(y = -x + 2\).
By plotting these lines on a coordinate plane, we can directly see the actual region that is being enclosed. This aids in verifying the calculated intersection points and understanding which portions of the lines form the boundary of the region.
To sketch these curves effectively:

• Identify the slope and intercept from each equation; for instance, \(y = x\) and \(y = 4x\) both pass through the origin but have different slopes.
• Observe that \(y = -x + 2\) is a line with a negative slope and a y-intercept at 2, altering the shape of the enclosed region.

By organizing and visualizing the curves via coordinate geometry, you gain a strong grasp of their layout and relationships, easing the application of calculus techniques to find the area.