Calculating Area With Definite Integrals The Function F(x) = 9 - X²

Hey everyone! Today, we're diving into a fascinating application of definite integrals: calculating the area between a curve and the x-axis. We'll be using the function f(x) = 9 - x² over the interval [0, 6] as our example. So, buckle up and let's get started!

Understanding the Problem: Area Between a Curve and the x-axis

So, what exactly does it mean to find the area between a curve and the x-axis? Imagine graphing our function, f(x) = 9 - x². You'll see a parabola opening downwards. The area we're interested in is the region enclosed between this curve and the x-axis within our specified interval, [0, 6].

But here's a crucial point: areas below the x-axis are considered negative when calculated using definite integrals directly. This is because the integral essentially sums up signed areas, where areas above the x-axis are positive and those below are negative. Therefore, to find the total area, we need to be mindful of where the function crosses the x-axis within our interval.

Breaking Down the Interval and Finding Intersections

Before we jump into the integration, let's figure out where our function, f(x) = 9 - x², intersects the x-axis within the interval [0, 6]. This is where f(x) = 0. So, we solve the equation:

9 - x² = 0 x² = 9 x = ±3

We find two intersections: x = 3 and x = -3. However, since our interval is [0, 6], we only care about the intersection at x = 3. This intersection divides our interval into two subintervals: [0, 3] and [3, 6].

On the interval [0, 3], the function f(x) = 9 - x² is above the x-axis, so the area will be positive. On the interval [3, 6], the function is below the x-axis, meaning the area will be negative. To get the total area, we'll need to calculate the areas on these subintervals separately and then take the absolute value of the area on [3, 6] before adding them together. This is crucial for obtaining the correct result.

Setting Up the Definite Integrals

Now that we've analyzed the behavior of our function and identified the subintervals, we can set up the definite integrals. We'll have two integrals, one for each subinterval:

• Area 1 (from x = 0 to x = 3): ∫[0, 3] (9 - x²) dx
• Area 2 (from x = 3 to x = 6): ∫[3, 6] (9 - x²) dx

Remember, we'll take the absolute value of the result from the second integral before adding it to the first to account for the area below the x-axis. This is a critical step to ensure we're calculating the total area between the curve and the x-axis.

Evaluating the Definite Integrals

Alright, guys, let's get our hands dirty and evaluate those definite integrals! We'll start by finding the antiderivative of our function, f(x) = 9 - x².

Finding the Antiderivative

The antiderivative of 9 is simply 9x. For -x², we use the power rule for integration, which states that the antiderivative of x^n is (x^(n+1))/(n+1). So, the antiderivative of -x² is (-x³)/3.

Therefore, the antiderivative of f(x) = 9 - x² is F(x) = 9x - (x³/3).

Evaluating the First Integral: ∫[0, 3] (9 - x²) dx

To evaluate the definite integral, we'll use the Fundamental Theorem of Calculus. This theorem tells us that the definite integral of a function from a to b is equal to the difference between the antiderivative evaluated at b and the antiderivative evaluated at a:

∫[a, b] f(x) dx = F(b) - F(a)

For our first integral, ∫[0, 3] (9 - x²) dx, we have:

F(3) = 9(3) - (3³/3) = 27 - 9 = 18 F(0) = 9(0) - (0³/3) = 0

So, the first area is:

Area 1 = F(3) - F(0) = 18 - 0 = 18

Evaluating the Second Integral: ∫[3, 6] (9 - x²) dx

Now, let's tackle the second integral, ∫[3, 6] (9 - x²) dx. We already have the antiderivative, F(x) = 9x - (x³/3), so we just need to evaluate it at the limits of integration:

F(6) = 9(6) - (6³/3) = 54 - 72 = -18 F(3) = 9(3) - (3³/3) = 27 - 9 = 18

So, the result of the second integral is:

∫[3, 6] (9 - x²) dx = F(6) - F(3) = -18 - 18 = -36

Remember, this area is negative because the function is below the x-axis on this interval. We need to take the absolute value to get the actual area:

|Area 2| = |-36| = 36

Calculating the Total Area

We're almost there! We've calculated the areas on both subintervals. Now, to find the total area between the curve and the x-axis, we simply add the absolute values of the areas:

Total Area = Area 1 + |Area 2| = 18 + 36 = 54

So, the total area between the curve f(x) = 9 - x² and the x-axis over the interval [0, 6] is 54 square units. This is the final answer we were seeking.

Summarizing the Steps for Success

Let's quickly recap the steps we took to solve this problem:

1. Understand the problem: Visualize the area between the curve and the x-axis.
2. Find x-intercepts: Determine where the function crosses the x-axis within the interval.
3. Divide the interval: Split the interval into subintervals based on the x-intercepts.
4. Set up the integrals: Write definite integrals for each subinterval.
5. Find the antiderivative: Calculate the antiderivative of the function.
6. Evaluate the integrals: Use the Fundamental Theorem of Calculus to evaluate each definite integral.
7. Take absolute values: Take the absolute value of any negative areas (areas below the x-axis).
8. Add the areas: Sum the absolute values of the areas from each subinterval to get the total area.

By following these steps carefully, you can confidently tackle any problem involving finding the area between a curve and the x-axis using definite integrals. Remember, practice makes perfect, so keep working on those problems, guys! You've got this!

Importance of Understanding the Concept

While we've successfully calculated the area for this specific function and interval, it's incredibly important to understand the underlying concept. This technique isn't just a mathematical trick; it's a powerful tool with applications in various fields.

For example, in physics, you might use this method to calculate the distance traveled by an object given its velocity function. In economics, it could be used to determine consumer surplus or producer surplus. Understanding the concept allows you to apply this technique to a wide range of problems, not just those that look exactly like our example.

Common Mistakes to Avoid

Before we wrap up, let's discuss some common mistakes people make when calculating the area between a curve and the x-axis. Avoiding these pitfalls will help you get the correct answer consistently.

• Forgetting to find x-intercepts: This is a big one! If you don't identify where the function crosses the x-axis, you won't be able to divide the interval correctly and you'll likely get the wrong answer.
• Not taking absolute values: As we discussed, areas below the x-axis are negative. If you don't take the absolute value of these areas before adding them to the positive areas, you'll underestimate the total area.
• Incorrectly evaluating the antiderivative: Double-check your antiderivative calculations. A small mistake here can throw off your entire answer.
• Misapplying the Fundamental Theorem of Calculus: Make sure you're subtracting the antiderivative evaluated at the lower limit of integration from the antiderivative evaluated at the upper limit. The order matters!

Practice Makes Perfect

The best way to master this concept is to practice, practice, practice! Work through different examples with varying functions and intervals. Pay close attention to the steps we've outlined, and be mindful of the common mistakes. The more you practice, the more comfortable you'll become with this technique, and the more confident you'll be in your ability to solve these problems. Remember, consistent effort is key to mastering any mathematical concept.

Conclusion: Mastering Area Calculation with Definite Integrals

Well, guys, we've reached the end of our journey into finding the area between a curve and the x-axis using definite integrals. We've covered a lot of ground, from understanding the basic concept to working through a detailed example and identifying common mistakes. Hopefully, you now have a solid understanding of this powerful technique.

Remember, the key to success is understanding the underlying principles and practicing diligently. Don't be afraid to tackle challenging problems, and always double-check your work. With perseverance, you'll master this concept and be able to apply it to a wide range of real-world applications.

Keep practicing, keep exploring, and most importantly, keep learning! You've got this!