Question

Sketch the following regions (if a figure is not given) and then find the area. The two regions in the first quadrant bounded by \(y=4 x-x^{2}\) \(y=4 x-4,\) and \(x=0\)

Short answer

Question: Sketch the regions in the first quadrant bounded by the functions \(y = 4x-x^2, y = 4x-4\), and \(x=0\), and find the area of those regions. Answer: The area of the regions bounded by the given functions in the first quadrant is \(\frac{16}{3} \text{ square units}\).

Step-by-step solution

Sketch the regions

The three functions are given as follows: 1. \(y = 4x - x^2\) 2. \(y = 4x - 4\) 3. \(x = 0\) (This is the y-axis) Sketch the regions on a coordinate plane. Draw dashed lines for any intersection points that you find.

Find Intersection Points

To determine the regions more precisely, we need to find the intersection points between the given functions. 1. Intersection point of \(y = 4x - x^2\) and \(y = 4x - 4\) Set the functions equal and solve for \(x\) to find the intersection points. \((4x - x^2) = (4x - 4)\) Solve for \(x\): \(x^2 - 4 = 0\) \(x = 2, -2\) Since our region is in the first quadrant, we will only consider the positive value of x i.e., \(x=2\). For this value of x, we find the value of y by inserting x into one of the functions: \(y = 4(2) - (2)^2 = 4\) Therefore, the intersection point is at \((2, 4)\). 2. Intersection point of \(y = 4x - x^2\) and \(x = 0\) Since \(x = 0\), we can simply plug in \(x=0\) into the first function to find y: \(y = 4(0) - (0)^2 = 0\) The intersection point is at \((0, 0)\). 3. Intersection point of \(y= 4x-4\) and \(x=0\) Similarly, plug in \(x=0\) into the second function: \(y=4(0)-4=-4\) As we're in the first quadrant we don't need to consider this intersection. Putting all the intersection points together, we have \((0, 0)\), and \((2, 4)\). Now, we will use these points as bounds for our definite integral to find the area of the regions.

Calculate the Area of the Regions

As we are in the first quadrant, we can integrate using the definite integral and take the difference between the two functions. The integral we need to compute must be in the form: \(Area = \int_{a}^{b} (f(x) - g(x)) dx\) In our case, we need to compute the following integral: \(Area = \int_{0}^{2} [(4x - x^2) - (4x - 4)] dx\) Let's simplify the integral expression: \(Area = \int_{0}^{2} (x^2 - 4) dx\) Now, find the antiderivative: \(F(x) = \frac{1}{3}x^3 - 4x\) Now, apply the Fundamental Theorem of Calculus and find the area of the region: \(Area = F(2) - F(0) = \frac{1}{3}(2)^3 - 4(2) - [\frac{1}{3}(0)^3 - 4(0)] = \frac{8}{3} - 8 = \frac{-16}{3}\) Since Area cannot be negative, we take the absolute value of the result. Therefore: \(Area = \frac{16}{3} \text{ square units}\) The area of the regions bounded by the given functions in the first quadrant is \(\frac{16}{3} \text{ square units}\).

Key concepts

Definite Integrals

One of the most important concepts in calculus is the definite integral. It helps us calculate the accumulated area under a curve, making it an essential tool for finding the area between curves as well. When solving problems involving definite integrals, we determine the limits of integration by finding the points of intersection, which we'll discuss further in the next sections. These points become our bounds as we integrate a function over a specific interval.

For instance, if we want to find the area between two curves from point \(a\) to \(b\), the definite integral formula we use is:

• \( \int_{a}^{b} (f(x) - g(x)) \, dx \)

Here, \(f(x)\) and \(g(x)\) represent the functions that enclose the region whose area we want to find. The integral calculates the difference between the two functions across the given interval, producing the area between them. This method is crucial when we're dealing with more complex regions bounded by multiple functions.

Area of Regions

In calculus, calculating the area of regions involves breaking down more complex shapes into manageable areas that we can compute using integrals. The challenge often lies in accurately identifying the boundaries of these regions, which is where sketching the curves and finding intersection points are most helpful. Once you've established these boundaries, finding the area becomes a process of integrating the difference between two functions on the interval dictated by the points you determined.

• The region's total area can be found by integrating the upper function minus the lower function.
• This process benefits significantly from understanding the geometric interpretations of the functions involved.

In the exercise, once we determine that \(y = 4x - x^2\) is the upper function and \(y = 4x - 4\) is the lower function between the intersection points, we compute the definite integral of their difference over the region of interest to find the area.

Intersection Points

Intersection points are critical when determining the areas of regions bounded by multiple functions. These points mark where functions intersect, creating boundaries for regions. When given two functions, finding their intersection points involves setting them equal and solving for \(x\). This process reveals where the curves meet on the coordinate plane, delimiting the area to be calculated.

• For the exercise, the intersection points of \(y = 4x - x^2\) and \(y = 4x - 4\) were found by equating the functions: \((4x - x^2) = (4x - 4)\).
• The solution \(x = 2\) (since \(x = -2\) doesn't fit the first quadrant condition) was used to calculate the corresponding \(y\)-value, helping define our region in the first quadrant.

These intersection points determine the limits of integration, which are essential for setting up the definite integral to find the area of the region.

Quadrants

In coordinate geometry, quadrants refer to the four divisions of the Cartesian plane. Each quadrant is a section of the plane where "x" and "y" have specific signs:

• Quadrant I: \((+,+)\) – both \(x\) and \(y\) are positive.
• Quadrant II: \((-,+)\)
• Quadrant III: \((-,-)\)
• Quadrant IV: \((+,-)\)

For our purposes, we focus only on the first quadrant, where both coordinate values are positive. The exercise specifies that the area lies in this region, guiding our considerations for which intersection points and function values to use when calculating area.

Staying within the first quadrant means any solutions or regions calculated must obey the condition \(x \ge 0\) and \(y \ge 0\). For instance, when determining where \(y = 4x - 4\) intersects with the line \(x = 0\), we neglect the result \((0, -4)\) since it does not lie within the first quadrant. Understanding quadrants allows us to restrict our solution to the meaningful positive parts of the plane and ensure calculations and interpretations are contextually correct.