Setting Up Definite Integral To Find Area Between Curves Y=18-x^2 And Y=-3x-52

How to set up the definite integral required to find the area of the region between the graph of $y=18-x^2$ and $y=-3x-52$ over the interval $-6 \leq x \leq 7$?

In calculus, finding the area between two curves is a common application of definite integrals. This article will guide you through the process of setting up the definite integral required to calculate the area of the region bounded by the graphs of two functions, $y = 18 - x^2$ and $y = -3x - 52$, over the interval $-6 \leq x \leq 7$. We'll explore the steps involved, including identifying the points of intersection, determining the upper and lower functions, and setting up the integral expression. Understanding these concepts is crucial for mastering integral calculus and its applications in various fields.

To begin, let's clearly define the problem. We are tasked with finding the area of the region enclosed between two curves: a parabola defined by the equation $y = 18 - x^2$ and a line defined by the equation $y = -3x - 52$. The interval of interest is $-6 \leq x \leq 7$, which means we are only concerned with the area bounded by these curves within this specific range of x-values. The key here is to visualize or sketch the graphs of these functions to understand how they intersect and which function lies above the other within the given interval.

Before we dive into the calculations, it’s important to grasp the underlying concept of using definite integrals to find areas. A definite integral essentially sums up infinitesimal rectangles under a curve (or between curves) to give the total area. The height of these rectangles is determined by the difference between the two functions, and the width is an infinitesimally small change in x, denoted as dx. Therefore, setting up the integral correctly involves identifying which function represents the "top" boundary and which represents the "bottom" boundary over the interval.

The first step in solving this type of problem is to find the points of intersection between the two curves. These points define the boundaries where the curves switch positions relative to each other (i.e., where the "top" function becomes the "bottom" function, or vice versa). Finding these points is crucial for setting up the correct limits of integration and ensuring accurate area calculation. We find these points by setting the two functions equal to each other and solving for x. This will give us the x-coordinates where the curves intersect. Next, we need to determine which function is greater than the other within the given interval. This can be done by either sketching the graphs or by choosing a test point within the interval and evaluating both functions at that point. The function with the higher value at the test point is the "top" function in that region. Once we know the top and bottom functions, we can set up the definite integral by subtracting the bottom function from the top function and integrating over the specified interval. This integral represents the area between the two curves. Finally, evaluating the definite integral will give us the numerical value of the area. This involves finding the antiderivative of the integrand (the difference between the functions) and applying the Fundamental Theorem of Calculus, which states that the definite integral is equal to the difference in the antiderivative evaluated at the upper and lower limits of integration.

The initial step in determining the area between the curves is to identify their points of intersection. These points define the boundaries where the curves may switch positions relative to each other, influencing which function lies above the other. To find these intersection points, we set the two functions equal to each other:

$$18 - x^2 = -3x - 52$$

Now, we rearrange the equation to form a quadratic equation:

$$x^2 - 3x - 70 = 0$$

This quadratic equation can be factored as follows:

$$(x - 10)(x + 7) = 0$$

Thus, the solutions for x are:

$$x = 10 \quad \text{and} \quad x = -7$$

These x-values represent the points where the two curves intersect. However, it's crucial to consider the given interval, $-6 \leq x \leq 7$. We observe that $x = -7$ falls outside this interval, so we discard it. The intersection point we are interested in is at $x = -7$, which is outside our interval of $-6 \leq x \leq 7$, and we have to consider $x = 10$ which is also outside the interval. However, we need to consider the interval endpoints as well since the area is bounded by the vertical lines at $x = -6$ and $x = 7$. So, within our interval of interest, the only relevant intersection point is x = -6 and x = 7, but we will verify the relative positions of the curves within the interval -6 <= x <= 7.

In the interval $-6 \leq x \leq 7$, we need to confirm if there is an intersection. As we calculated, the two points of intersection are outside the interval, which means there is no intersection within the interval. This means that one function is consistently above the other throughout the interval. This simplifies the problem because we don't need to divide the interval into subintervals based on changes in the relative positions of the curves. However, it is important to note that this scenario highlights the necessity of always comparing the intersection points with the boundaries defined in the problem statement. Neglecting this step could lead to incorrect integration limits and, consequently, an incorrect calculation of the area between the curves.

With the intersection points found (or in this case, the lack of intersection points within the interval), the next crucial step is to determine which function lies above the other within the interval $-6 \leq x \leq 7$. This is essential for setting up the definite integral correctly, as we need to subtract the lower function from the upper function to find the area between the curves. There are two primary methods to achieve this: sketching the graphs or using a test point.

1. Sketching the Graphs:

Sketching the graphs of $y = 18 - x^2$ and $y = -3x - 52$ provides a visual representation of their relative positions. The graph of $y = 18 - x^2$ is a downward-opening parabola with its vertex at (0, 18). The graph of $y = -3x - 52$ is a straight line with a negative slope. By sketching these graphs, we can visually determine which function has higher y-values within the interval $-6 \leq x \leq 7$. The parabola will be above the line within this interval.

2. Using a Test Point:

Alternatively, we can choose a test point within the interval $-6 \leq x \leq 7$ and evaluate both functions at that point. For simplicity, let's choose $x = 0$ as our test point. Evaluating the functions at $x = 0$ gives us:

• $y = 18 - (0)^2 = 18$
• $y = -3(0) - 52 = -52$

Since 18 is greater than -52, the function $y = 18 - x^2$ lies above the function $y = -3x - 52$ at $x = 0$. Since there are no intersection points within the interval, this relationship holds true for the entire interval $-6 \leq x \leq 7$. This confirms that the parabola is above the line throughout the interval, which is crucial information for correctly setting up the definite integral.

In conclusion, we've determined that $y = 18 - x^2$ is the upper function and $y = -3x - 52$ is the lower function within the interval $-6 \leq x \leq 7$. This finding is the cornerstone for the next step, which involves setting up the definite integral to calculate the area between these two curves. The ability to accurately identify the upper and lower functions is paramount in this type of problem, as it directly impacts the sign and, consequently, the value of the calculated area. A misunderstanding here can lead to incorrect results, highlighting the importance of careful analysis and visualization techniques like sketching graphs or employing the test point method.

Now that we've identified the upper function ($y = 18 - x^2$) and the lower function ($y = -3x - 52$) within the interval $-6 \leq x \leq 7$, we can set up the definite integral to calculate the area between the curves. The fundamental principle here is that the area between two curves is given by the integral of the difference between the upper function and the lower function, integrated over the given interval.

The general formula for the area (A) between two curves, $f(x)$ and $g(x)$, over the interval $[a, b]$, where $f(x) \geq g(x)$ on $[a, b]$, is:

$$A = \int_{a}^{b} [f(x) - g(x)] dx$$

In our specific case, $f(x) = 18 - x^2$, $g(x) = -3x - 52$, $a = -6$, and $b = 7$. Plugging these values into the formula, we get:

$$A = \int_{-6}^{7} [(18 - x^2) - (-3x - 52)] dx$$

This integral represents the area between the parabola $y = 18 - x^2$ and the line $y = -3x - 52$ from $x = -6$ to $x = 7$. Before evaluating the integral, it's often helpful to simplify the integrand:

$$A = \int_{-6}^{7} (18 - x^2 + 3x + 52) dx$$

$$A = \int_{-6}^{7} (-x^2 + 3x + 70) dx$$

This simplified integral is now ready for evaluation. The setup of the definite integral is a critical step in solving area problems. It encapsulates the geometric concept of summing infinitesimal rectangles between the curves and translates it into a mathematical expression that can be solved using calculus techniques. The integrand, which represents the height of these rectangles, is the difference between the upper and lower functions. The limits of integration define the interval over which we are summing these rectangles. A correctly set up integral ensures that we are accurately capturing the area of the desired region.

In this comprehensive exploration, we have successfully set up the definite integral required to find the area of the region bounded by the graphs of $y = 18 - x^2$ and $y = -3x - 52$ over the interval $-6 \leq x \leq 7$. We meticulously walked through each step, from identifying the points of intersection (or the absence thereof within the interval) to determining the upper and lower functions, and finally, constructing the definite integral expression. The integral we've established is:

$$A = \int_{-6}^{7} (-x^2 + 3x + 70) dx$$

This integral represents the mathematical formulation necessary to calculate the area. While we haven't evaluated the integral in this article, the setup is the crucial foundation for finding the numerical value of the area. To complete the process, one would proceed by finding the antiderivative of the integrand and then applying the Fundamental Theorem of Calculus to evaluate the integral at the limits of integration.

This exercise underscores the importance of a systematic approach to solving area problems in calculus. It highlights the need to understand the underlying concepts, visualize the curves, and carefully consider the interval of interest. The ability to accurately set up the definite integral is a fundamental skill in calculus and has wide-ranging applications in various fields, including physics, engineering, and economics. The process outlined in this article serves as a valuable guide for tackling similar problems and further solidifying one's understanding of integral calculus.