Enclosed Area Between Curves: A Calculus Guide

by Mei Lin 47 views

Hey guys! Today, we're diving into a fun problem from calculus: finding the area enclosed by the graphs of two functions. Specifically, we'll be working with the functions y = 2x² + 11 and y = -4x + 17. This is a classic problem that combines algebra and integral calculus, so let's break it down step by step.

1. Understanding the Problem

Before we jump into calculations, let's visualize what we're trying to find. We have two functions:

  • A parabola: y = 2x² + 11
  • A straight line: y = -4x + 17

These two graphs will intersect at certain points, and the area enclosed between them is what we need to determine. Think of it as the space trapped between the curves. To accurately find the enclosed area, it’s essential to first understand the behavior of each function. The parabola y = 2x² + 11 opens upwards and has a minimum value at y = 11, while the line y = -4x + 17 has a negative slope and intersects the y-axis at 17. Visualizing these functions helps in anticipating the region we will be calculating.

Imagine plotting these two equations on a graph. The parabola, with its U-shape, sits above the x-axis, and the line cuts across it. The points where the line intersects the parabola will define the boundaries of our enclosed area. Our mission is to calculate this area precisely. This involves several key steps, starting with finding the points of intersection, setting up the integral, and finally, evaluating the integral. It’s a journey through algebra and calculus, so buckle up and let’s dive in! Remember, this isn’t just about finding a numerical answer; it’s about understanding the process and the underlying mathematical concepts. The ability to find the enclosed area between curves is a fundamental skill in calculus with applications in various fields, from physics to economics.

2. Finding the Points of Intersection

The first crucial step is to find the points of intersection. These points define the limits of our integration. To find them, we need to set the two equations equal to each other:

  • 2x² + 11 = -4x + 17

Now, let's solve this quadratic equation. First, rearrange the terms to get a standard quadratic form:

  • 2x² + 4x - 6 = 0

We can simplify this by dividing the entire equation by 2:

  • x² + 2x - 3 = 0

Now, we can factor this quadratic equation:

  • (x + 3)(x - 1) = 0

This gives us two solutions for x:

  • x = -3
  • x = 1

These are the x-coordinates of our intersection points. To find the corresponding y-coordinates, we can plug these x-values back into either of the original equations. Let's use the linear equation y = -4x + 17:

For x = -3:

  • y = -4(-3) + 17 = 12 + 17 = 29

For x = 1:

  • y = -4(1) + 17 = -4 + 17 = 13

So, our points of intersection are (-3, 29) and (1, 13). These points are super important because they tell us where the line and the parabola meet, and they will serve as the boundaries for our integral calculation. Think of these points as the anchors that define the area we want to measure. We now know that we need to find the enclosed area between x = -3 and x = 1. Understanding how to find the points of intersection is a fundamental skill in calculus and is essential for solving a variety of problems involving curves and graphs.

3. Setting up the Integral

Now that we have the points of intersection, we can set up the integral to calculate the area. The key idea here is that the area between two curves, f(x) and g(x), from x = a to x = b is given by:

  • ∫[a to b] |f(x) - g(x)| dx

Where f(x) is the upper function and g(x) is the lower function within the interval [a, b]. In our case, we need to determine which function is on top and which is on the bottom within the interval [-3, 1].

Let's consider a value between -3 and 1, say x = 0. Plug this into both equations:

  • y = 2(0)² + 11 = 11
  • y = -4(0) + 17 = 17

At x = 0, the line y = -4x + 17 is above the parabola y = 2x² + 11. This suggests that the line is the upper function within our interval.

Therefore, our integral will be:

  • ∫[-3 to 1] ((-4x + 17) - (2x² + 11)) dx

Let's simplify the integrand:

  • ∫[-3 to 1] (-2x² - 4x + 6) dx

This integral represents the area enclosed between the two curves. Setting up the integral correctly is crucial; it's the foundation upon which our calculation rests. We’ve identified the limits of integration (-3 and 1) and determined the order of the functions within the integral. This step demonstrates a deep understanding of the relationship between the functions and the area they enclose. Now, we’re ready to find the enclosed area by evaluating this definite integral. Remember, the ability to find the enclosed area by setting up the integral is a cornerstone of integral calculus.

4. Evaluating the Integral

Now comes the fun part: evaluating the integral! We have:

  • ∫[-3 to 1] (-2x² - 4x + 6) dx

To evaluate this, we'll find the antiderivative of the integrand and then apply the Fundamental Theorem of Calculus.

The antiderivative of -2x² is (-2/3)x³. The antiderivative of -4x is -2x². The antiderivative of 6 is 6x.

So, the antiderivative of -2x² - 4x + 6 is:

  • (-2/3)x³ - 2x² + 6x

Now, we'll evaluate this antiderivative at the limits of integration, 1 and -3, and subtract the results:

[((-2/3)(1)³ - 2(1)² + 6(1)) - ((-2/3)(-3)³ - 2(-3)² + 6(-3))]

Let's simplify this:

[(-2/3 - 2 + 6) - ((-2/3)(-27) - 2(9) - 18)]

[(-2/3 + 4) - (18 - 18 - 18)]

[(10/3) - (-18)]

(10/3) + 18

(10/3) + (54/3)

64/3

So, the area enclosed by the graphs of the functions is 64/3 square units. This is our final answer! Evaluating the integral involves applying the power rule for integration and carefully substituting the limits of integration. It’s a process that highlights the power of calculus in solving geometric problems. We’ve successfully found the enclosed area by breaking down the integral into manageable steps. This calculation underscores the importance of precision and attention to detail in calculus. The final result, 64/3 square units, represents the exact area trapped between the parabola and the line. Remember, being able to find the enclosed area through integral evaluation is a vital skill in calculus.

5. Conclusion

Woohoo! We made it! We successfully found the enclosed area between the graphs of y = 2x² + 11 and y = -4x + 17. We started by understanding the problem, then we found the points of intersection, set up the integral, and finally, evaluated it. The area enclosed is 64/3 square units.

This problem demonstrates the power of calculus in finding areas bounded by curves. It combines algebraic techniques for solving equations with the fundamental concepts of integral calculus. Remember, the key steps are:

  1. Find the points of intersection to determine the limits of integration.
  2. Determine which function is the upper function and which is the lower function within the interval.
  3. Set up the integral correctly, subtracting the lower function from the upper function.
  4. Evaluate the integral using the Fundamental Theorem of Calculus.

Calculus can seem daunting at first, but by breaking it down into smaller, manageable steps, you can conquer even the most challenging problems. Keep practicing, and you'll become a calculus pro in no time! This journey of finding the enclosed area has shown us the beauty and applicability of calculus. By mastering these techniques, you’ll be well-equipped to tackle a wide range of problems in mathematics, physics, engineering, and beyond. So keep up the great work, and remember, the world of calculus is vast and exciting, full of opportunities to explore and discover. Understanding how to find the enclosed area is just the beginning of your calculus adventure!