Area Between Curve And X-axis: Y=x^2+4x From X=-2 To 0

Hey guys! Today, we're diving into a fascinating topic in calculus: calculating the area between a curve and the x-axis. Specifically, we'll tackle the problem of finding the area between the curve y = x² + 4x and the x-axis from x = -2 to x = 0. This might sound intimidating at first, but don't worry, we'll break it down step by step. Let's get started!

Understanding the Problem

Before we jump into the calculations, let's make sure we understand what we're trying to achieve. The core concept here involves using definite integrals to determine the area. Think of it this way: we're essentially summing up an infinite number of infinitely thin rectangles under the curve between our specified limits (x = -2 and x = 0). So, the key to finding the area between a curve and the x-axis lies in the realm of integral calculus, where we use definite integrals to calculate the accumulated quantity, in this case, the area, within specified bounds.

The curve we're dealing with is a parabola, defined by the equation y = x² + 4x. To visualize this, imagine a U-shaped curve on a graph. The x-axis, as we all know, is the horizontal line where y = 0. We're interested in the area enclosed between this parabola and the x-axis within the vertical lines x = -2 and x = 0. It's super important to note that parts of the curve might dip below the x-axis in this interval. When this happens, the function values (y-values) are negative, and the definite integral will give us a negative value for the area. This might seem counterintuitive, as area is typically considered a positive quantity. However, this negative sign is crucial because it tells us that the area lies below the x-axis. To get the total area, we will have to consider the absolute value of any such negative results.

Understanding the definite integral is paramount. The definite integral of a function f(x) from a to b, denoted as ∫[a to b] f(x) dx, represents the signed area between the curve y = f(x) and the x-axis from x = a to x = b. The term "signed area" is key here because it accounts for the portions of the curve that lie below the x-axis, which contribute negatively to the total integral value. The limits of integration, a and b, define the interval over which we are calculating the area. In our specific problem, f(x) = x² + 4x, a = -2, and b = 0. So, setting up the definite integral correctly is the first crucial step in solving this problem.

Setting Up the Integral

Okay, let's get down to the nitty-gritty. The first step is to set up the definite integral. Remember, the area between the curve y = f(x) and the x-axis from x = a to x = b is given by the definite integral:

∫[a to b] f(x) dx

In our case, f(x) = x² + 4x, a = -2, and b = 0. So, our integral looks like this:

∫[-2 to 0] (x² + 4x) dx

This integral represents the signed area between the curve y = x² + 4x and the x-axis from x = -2 to x = 0. Before we actually calculate the integral, it's a good idea to visualize the function within the given interval. This can help us anticipate whether the area will be positive or negative and give us a better understanding of the problem.

Now, before we proceed, it's wise to analyze the function f(x) = x² + 4x in the interval [-2, 0]. This involves determining where the function is positive, negative, or zero within this interval. To do this, we first find the roots of the equation x² + 4x = 0. Factoring, we get x(x + 4) = 0, which gives us roots at x = 0 and x = -4. Only x = 0 lies within our interval of interest, but x = -4 helps us understand the overall behavior of the parabola. Since the parabola opens upwards (the coefficient of x² is positive), it will be negative between the roots and positive outside the roots. Thus, within the interval (-2, 0), the function x² + 4x is negative. This is a crucial piece of information, as it tells us that the definite integral will yield a negative value, representing an area below the x-axis. To find the actual area, we'll need to take the absolute value of the result.

The importance of correctly setting up the integral cannot be overstated. It's the foundation upon which the rest of the solution is built. A mistake here will propagate through the entire calculation, leading to an incorrect answer. Understanding the function and the interval of integration allows us to anticipate the sign of the result and helps us interpret the final answer correctly. This step-by-step approach ensures we are clear on every aspect of the problem before we proceed to the calculation phase.

Evaluating the Integral

Alright, let's get our hands dirty and evaluate the integral! To do this, we'll use the power rule for integration, which states that:

∫xⁿ dx = (x^(n+1)) / (n+1) + C

where C is the constant of integration. However, since we're dealing with a definite integral, we don't need to worry about the constant C because it will cancel out when we evaluate the integral at the limits of integration.

Let's apply this to our integral:

∫[-2 to 0] (x² + 4x) dx

First, we find the antiderivative of x² + 4x:

∫x² dx = (x³)/3

∫4x dx = 4 * (x²)/2 = 2x²

So, the antiderivative of x² + 4x is (x³)/3 + 2x².

Now, we need to evaluate this antiderivative at the limits of integration, x = 0 and x = -2, and subtract the result at the lower limit from the result at the upper limit. This is known as the Fundamental Theorem of Calculus.

Evaluating at x = 0:

((0)³)/3 + 2(0)² = 0

Evaluating at x = -2:

((-2)³)/3 + 2(-2)² = (-8)/3 + 2(4) = (-8)/3 + 8 = (-8 + 24)/3 = 16/3

Now, we subtract the value at the lower limit from the value at the upper limit:

0 - (16/3) = -16/3

So, the value of the definite integral is -16/3. Remember that this is a negative value, which indicates that the area lies below the x-axis.

The power rule of integration is a fundamental tool in calculus, and mastering its application is crucial for solving a wide range of problems. In our case, applying the power rule to x² and 4x allows us to find their respective antiderivatives, which are essential for evaluating the definite integral. The antiderivative represents the reverse process of differentiation, and it gives us a function whose derivative is the original function we are integrating. This step-by-step process of finding the antiderivative and then evaluating it at the limits of integration is the heart of the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus is the cornerstone of integral calculus, linking differentiation and integration in a profound way. It tells us that to evaluate a definite integral, we simply need to find the antiderivative of the integrand, evaluate it at the upper and lower limits of integration, and subtract the results. This theorem provides a powerful and efficient method for calculating definite integrals, which are essential for finding areas, volumes, and other accumulated quantities. In our problem, the application of this theorem allows us to move from the abstract representation of the integral to a concrete numerical value representing the signed area.

Finding the Area

We've calculated the value of the definite integral to be -16/3. However, as we discussed earlier, this is a signed area. Since we're interested in the actual area, which is always a positive quantity, we need to take the absolute value of our result.

| -16/3 | = 16/3

Therefore, the area between the curve y = x² + 4x and the x-axis from x = -2 to x = 0 is 16/3 square units. We can also express this as a mixed number, which is 5 1/3 square units, or as a decimal, approximately 5.33 square units.

The absolute value is crucial here because area, in its conventional sense, is always a non-negative quantity. The definite integral, on the other hand, gives us the signed area, which takes into account the regions where the function lies below the x-axis. By taking the absolute value, we are essentially considering the magnitude of the area without regard to its sign, giving us the true geometric area. This step highlights the importance of understanding the relationship between the definite integral and the geometric interpretation of area.

The final answer, 16/3 square units, represents the total area enclosed between the curve and the x-axis within the specified interval. This area is a measure of the region bounded by the curve, the x-axis, and the vertical lines at x = -2 and x = 0. It's a tangible representation of the accumulation of the function's values over that interval. The ability to calculate such areas has numerous applications in various fields, including physics, engineering, and economics.

Conclusion

So, there you have it! We've successfully found the area between the curve y = x² + 4x and the x-axis from x = -2 to x = 0. We did this by setting up a definite integral, evaluating it using the power rule for integration and the Fundamental Theorem of Calculus, and then taking the absolute value to get the actual area. This problem perfectly illustrates how we can use calculus to solve real-world geometric problems. Keep practicing, and you'll become a pro at calculating areas under curves in no time!

Remember, the key takeaways are:

• Understand the concept of definite integrals and how they relate to area.
• Set up the integral correctly by identifying the function and the limits of integration.
• Evaluate the integral using appropriate techniques like the power rule.
• Take the absolute value of the result if you're looking for the actual area.

Keep exploring the fascinating world of calculus, guys! There's so much more to discover! This journey is not just about solving problems but about developing a deeper understanding of mathematical concepts and their applications.