Area Calculation Between Curves Solving Y=√x, Y=x-2, And Y=0

Finding the area enclosed by several functions is a common problem in calculus. This article aims to demonstrate a step-by-step approach to solve such problems. The specific problem we will tackle involves finding the area bounded by the functions $y = \sqrt{x}$, $y = x - 2$, and $y = 0$. This process combines algebraic manipulation, calculus techniques, and geometric interpretation to arrive at the solution. Understanding these concepts is crucial for various applications in physics, engineering, and economics, where calculating areas between curves is essential for modeling real-world scenarios and optimizing system performance.

1. Understanding the Problem and the Functions

Before diving into calculations, it’s crucial to understand the functions and the region we’re dealing with. We have three functions:

1. $y = \sqrt{x}$: This is the square root function, a curve that starts at the origin (0,0) and increases as x increases. It lies in the first quadrant where both x and y are non-negative. The graph is a smooth curve that grows more slowly as x gets larger.
2. $y = x - 2$: This is a linear function, representing a straight line with a slope of 1 and a y-intercept of -2. It intersects the x-axis at x = 2, which is an important point for defining the region.
3. $y = 0$: This is the x-axis, serving as one of the boundaries for the area we want to calculate. It’s a horizontal line passing through the origin and is fundamental in defining regions in the coordinate plane.

The key to finding the area is to first visualize or sketch these functions to understand the enclosed region. By plotting these functions, we can identify the points of intersection and the relevant boundaries. These points are crucial because they define the limits of integration, which are necessary for calculating the area. A careful sketch can prevent errors and provide a clear picture of the problem.

2. Finding the Intersection Points

To determine the area bounded by these curves, we need to find where they intersect. These intersection points will define the limits of integration. Let's find the intersection points of each pair of functions:

Intersection of $y = \sqrt{x}$ and $y = x - 2$

To find where these two functions intersect, we set them equal to each other:

$\sqrt{x} = x - 2$

To solve this equation, we square both sides to eliminate the square root:

$x = (x - 2)^2$ $x = x^2 - 4x + 4$

Rearrange the equation to form a quadratic equation:

$x^2 - 5x + 4 = 0$

Now, we factor the quadratic equation:

$(x - 4)(x - 1) = 0$

This gives us two potential solutions for x: x = 4 and x = 1. We need to check these solutions in the original equation $\sqrt{x} = x - 2$ because squaring both sides can introduce extraneous solutions.

For x = 4:

$\sqrt{4} = 4 - 2$ $2 = 2$ (This solution is valid)

For x = 1:

$\sqrt{1} = 1 - 2$ $1 = -1$ (This solution is extraneous)

Thus, the only valid intersection point is at x = 4. The corresponding y-coordinate is:

$y = \sqrt{4} = 2$

So, the intersection point is (4, 2).

Intersection of $y = \sqrt{x}$ and $y = 0$

Setting $y = \sqrt{x}$ equal to 0:

$\sqrt{x} = 0$

Squaring both sides, we get:

$x = 0$

Thus, the intersection point is (0, 0).

Intersection of $y = x - 2$ and $y = 0$

Setting $y = x - 2$ equal to 0:

$x - 2 = 0$

Solving for x:

$x = 2$

Thus, the intersection point is (2, 0).

Summary of Intersection Points

• $y = \sqrt{x}$ and $y = x - 2$ intersect at (4, 2).
• $y = \sqrt{x}$ and $y = 0$ intersect at (0, 0).
• $y = x - 2$ and $y = 0$ intersect at (2, 0).

These intersection points define the boundaries of the region we are interested in. Specifically, they show that the region is bounded by the x-axis (y = 0), the curve $y = \sqrt{x}$, and the line $y = x - 2$. The intersection points (0, 0), (2, 0), and (4, 2) will be crucial for setting up our integrals to calculate the area.

3. Setting up the Integrals

Now that we have the intersection points, we can set up the integrals to calculate the area. The region is bounded by the curves $y = \sqrt{x}$, $y = x - 2$, and $y = 0$. We need to divide the region into sections where the bounding functions are consistent.

Region Division

1. From x = 0 to x = 2: In this region, the area is bounded above by $y = \sqrt{x}$ and below by $y = 0$ (the x-axis).
2. From x = 2 to x = 4: In this region, the area is bounded above by $y = \sqrt{x}$ and below by $y = x - 2$.

Integral Setup

We will calculate the area of each region separately and then add them together.

1. Area 1 (from x = 0 to x = 2):

The area is given by the integral of the difference between the upper function ($y = \sqrt{x}$) and the lower function ($y = 0$) with respect to x:

$A_1 = \int_{0}^{2} (\sqrt{x} - 0) dx = \int_{0}^{2} \sqrt{x} dx$

2. Area 2 (from x = 2 to x = 4):

The area is given by the integral of the difference between the upper function ($y = \sqrt{x}$) and the lower function ($y = x - 2$) with respect to x:

$A_2 = \int_{2}^{4} (\sqrt{x} - (x - 2)) dx = \int_{2}^{4} (\sqrt{x} - x + 2) dx$

Total Area

The total area A is the sum of the two areas:

$A = A_1 + A_2 = \int_{0}^{2} \sqrt{x} dx + \int_{2}^{4} (\sqrt{x} - x + 2) dx$

This setup allows us to break down a complex area calculation into smaller, manageable integrals. Each integral represents the area between two curves over a specific interval, making it easier to compute the overall area. The precise limits of integration (0, 2, and 4) are critical for accurately capturing the bounded region's area. The next step involves evaluating these integrals, which will provide the numerical value of the area.

4. Evaluating the Integrals

Now we need to evaluate the integrals we set up in the previous section to find the area bounded by the curves. We have two integrals to compute:

$A_1 = \int_{0}^{2} \sqrt{x} dx$ $A_2 = \int_{2}^{4} (\sqrt{x} - x + 2) dx$

Evaluating $A_1$

First, we rewrite $\sqrt{x}$ as $x^{1/2}$ and find its antiderivative:

$\int x^{1/2} dx = \frac{x^{(1/2) + 1}}{(1/2) + 1} + C = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$

Now, we evaluate the definite integral:

$A_1 = \int_{0}^{2} \sqrt{x} dx = \left[\frac{2}{3}x^{3/2}\right]_{0}^{2} = \frac{2}{3}(2^{3/2}) - \frac{2}{3}(0^{3/2})$

$A_1 = \frac{2}{3}(2\sqrt{2}) = \frac{4\sqrt{2}}{3}$

Evaluating $A_2$

We need to find the antiderivative of each term in the integral:

$\int (\sqrt{x} - x + 2) dx = \int x^{1/2} dx - \int x dx + \int 2 dx$

We already know the antiderivative of $x^{1/2}$ is $\frac{2}{3}x^{3/2}$. The antiderivative of x is $\frac{1}{2}x^2$, and the antiderivative of 2 is 2x. So,

$\int (\sqrt{x} - x + 2) dx = \frac{2}{3}x^{3/2} - \frac{1}{2}x^2 + 2x + C$

Now, we evaluate the definite integral:

$A_2 = \int_{2}^{4} (\sqrt{x} - x + 2) dx = \left[\frac{2}{3}x^{3/2} - \frac{1}{2}x^2 + 2x\right]_{2}^{4}$

$A_2 = \left(\frac{2}{3}(4^{3/2}) - \frac{1}{2}(4)^2 + 2(4)\right) - \left(\frac{2}{3}(2^{3/2}) - \frac{1}{2}(2)^2 + 2(2)\right)$

$A_2 = \left(\frac{2}{3}(8) - 8 + 8\right) - \left(\frac{2}{3}(2\sqrt{2}) - 2 + 4\right)$

$A_2 = \frac{16}{3} - \frac{4\sqrt{2}}{3} - 2$

$A_2 = \frac{16}{3} - \frac{4\sqrt{2}}{3} - \frac{6}{3} = \frac{10}{3} - \frac{4\sqrt{2}}{3}$

Total Area Calculation

Now we add the two areas:

$A = A_1 + A_2 = \frac{4\sqrt{2}}{3} + \frac{10}{3} - \frac{4\sqrt{2}}{3}$

$A = \frac{10}{3}$

Numerical Approximation

To get a numerical approximation, we calculate:

$A = \frac{10}{3} \approx 3.333$

Thus, the area bounded by the curves $y = \sqrt{x}$, $y = x - 2$, and $y = 0$ is approximately 3.333 square units. This result provides a clear quantitative measure of the region's size, essential for practical applications in various scientific and engineering contexts.

5. Conclusion

In conclusion, we found the area bounded by the functions $y = \sqrt{x}$, $y = x - 2$, and $y = 0$ to be $\frac{10}{3}$ square units, which is approximately 3.333 square units when rounded to the nearest thousandth. This process involved several key steps:

1. Understanding the Functions: We started by analyzing the given functions and visualizing the region they enclose.
2. Finding Intersection Points: We determined the points where the functions intersect, which defined the limits of integration.
3. Setting up the Integrals: We divided the region into sections and set up definite integrals to calculate the area of each section.
4. Evaluating the Integrals: We computed the definite integrals to find the numerical values of the areas.
5. Total Area Calculation: We added the areas of the sections to find the total area.

This problem illustrates the fundamental techniques of calculus used to find areas between curves. The combination of algebraic manipulation, integral calculus, and geometric interpretation is crucial for solving such problems. Mastering these techniques is essential for more advanced topics in mathematics and their applications in real-world scenarios.

This example serves as a comprehensive guide to tackling similar area-finding problems. By systematically following these steps, one can accurately determine the area bounded by various functions, providing valuable insights for mathematical and practical applications.