Area Between Curve And Line: A Calculus Guide

Hey guys! Today, we're diving into a classic calculus problem: finding the area enclosed between the curve ${ y = x^2 - 2x - 3 }$ and the line ${ y = x + 1 }$. This is a super common type of question you'll see in calculus courses, and mastering it will definitely boost your confidence. So, let's break it down step by step.

1. Understanding the Problem

Before we jump into calculations, let's visualize what we're dealing with. We have a parabola ${ y = x^2 - 2x - 3 }$ and a straight line ${ y = x + 1 }$. The area we want to find is the region trapped between these two graphs. To find this area, we'll need to use integral calculus. The basic idea is to integrate the difference between the two functions over the interval where they intersect.

Key Concepts Recap:

• Area Between Curves: The area between two curves ${ f(x) }$ and ${ g(x) }$ from ${ x = a }$ to ${ x = b }$ is given by ${ \int_{a}^{b} |f(x) - g(x)| dx }$. We use the absolute value because we want the area to be positive, regardless of which function is on top.
• Intersection Points: Finding where the curves intersect is crucial because these points define the limits of integration (i.e., ${ a }$ and ${ b }$).

Why is this important? Think about it this way: integration is essentially summing up infinitely thin rectangles. When finding the area between curves, the height of each rectangle is the difference between the y-values of the two functions at a particular x-value. If we don't know where the curves intersect, we don't know where to start and stop summing these rectangles!

2. Finding the Intersection Points

The first thing we need to do is figure out where the curve and the line intersect. This will give us the limits of integration. To find the intersection points, we set the two equations equal to each other: ${ x^2 - 2x - 3 = x + 1 }$ Now, let's solve for ${ x }$: ${ x^2 - 2x - 3 - x - 1 = 0 }$ ${ x^2 - 3x - 4 = 0 }$ This is a quadratic equation, which we can factor: ${ (x - 4)(x + 1) = 0 }$ So, the solutions are ${ x = 4 }$ and ${ x = -1 }$. These are the x-coordinates of the intersection points. To find the corresponding y-coordinates, we can plug these x-values into either equation (the line equation is simpler, so let's use that):

• For ${ x = 4 }$: ${ y = 4 + 1 = 5 }$
• For ${ x = -1 }$: ${ y = -1 + 1 = 0 }$

Therefore, the intersection points are ${ (4, 5) }$ and ${ (-1, 0) }$.

Why bother finding intersection points? Imagine trying to calculate the area without knowing where the curves cross. You'd be integrating over a region that isn't actually enclosed by both curves, leading to a completely wrong answer. Finding the intersection points defines the precise boundaries of the area we're interested in.

3. Setting Up the Integral

Now that we have the intersection points, we know that we need to integrate from ${ x = -1 }$ to ${ x = 4 }$. The next step is to determine which function is "on top" in this interval. In other words, which function has larger y-values between ${ x = -1 }$ and ${ x = 4 }$?

We can pick a test point within the interval, say ${ x = 0 }$, and plug it into both equations:

• For the curve: ${ y = (0)^2 - 2(0) - 3 = -3 }$
• For the line: ${ y = 0 + 1 = 1 }$

Since ${ 1 > -3 }$, the line ${ y = x + 1 }$ is above the curve ${ y = x^2 - 2x - 3 }$ in the interval ${ [-1, 4] }$. Therefore, we'll subtract the curve's equation from the line's equation inside the integral.

The integral representing the area is: ${ \text{Area} = \int_{-1}^{4} [(x + 1) - (x^2 - 2x - 3)] dx }$

Why does "which function is on top" matter? If you subtract the functions in the wrong order, you'll get a negative value for the area. While the magnitude of the number would be correct, area is a positive quantity. Using the correct order ensures we get a positive result, representing the actual area enclosed.

4. Evaluating the Integral

Let's simplify the integrand and then evaluate the integral:

${ \text{Area} = \int_{-1}^{4} (x + 1 - x^2 + 2x + 3) dx }$

${ \text{Area} = \int_{-1}^{4} (-x^2 + 3x + 4) dx }$

Now, we find the antiderivative:

${ \text{Area} = \left[ -\frac{1}{3}x^3 + \frac{3}{2}x^2 + 4x \right]_{-1}^{4} }$

Next, we evaluate the antiderivative at the limits of integration:

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

${ \text{Area} = \left( -\frac{64}{3} + 24 + 16 \right) - \left( \frac{1}{3} + \frac{3}{2} - 4 \right) }$

${ \text{Area} = \left( -\frac{64}{3} + 40 \right) - \left( \frac{2}{6} + \frac{9}{6} - \frac{24}{6} \right) }$

${ \text{Area} = \left( \frac{-64 + 120}{3} \right) - \left( \frac{-13}{6} \right) }$

${ \text{Area} = \frac{56}{3} + \frac{13}{6} }$

${ \text{Area} = \frac{112}{6} + \frac{13}{6} }$

${ \text{Area} = \frac{125}{6} }$

So, the area enclosed between the curve and the line is ${ \frac{125}{6} }$ square units.

5. Conclusion

And there you have it! We successfully found the area enclosed between the parabola ${ y = x^2 - 2x - 3 }$ and the line ${ y = x + 1 }$. Remember, the key steps are:

1. Find the intersection points: Set the equations equal to each other and solve for ${ x }$.
2. Determine which function is on top: Pick a test point within the interval and compare the y-values.
3. Set up the integral: Integrate the difference between the top function and the bottom function, from the lower limit of integration to the upper limit.
4. Evaluate the integral: Find the antiderivative and evaluate it at the limits of integration.

This type of problem is a fundamental concept in calculus, and mastering it will definitely help you tackle more complex problems in the future. Keep practicing, and you'll become a pro in no time!

Further Practice:

Try varying the curve and line equations. What happens if you change the coefficients? What if you use a different type of curve, like a cubic function? Experimenting with different equations will help you solidify your understanding of the process.