Problem 41 - Precalculus with Limits

Question

In Exercises 37-48, use the limit process to find the area of the region between the graph of the function and the x-axis over the specified interval. $$ f(x) = 2 - x^2 $$ Interval \( [-1, 1] \)

Step-by-Step Solution

Answer

The area under the curve \(f(x) = 2 - x^2\) within the interval [-1,1] is \(2/3\) square units.

Step 1: Understand the function and the interval

We have the function \( f(x) = 2 - x^2 \) and the interval [-1, 1]. The function is a downward opening parabola, and the area to be calculated lies between the curve and the x-axis from x=-1 to x=1.

Step 2: Setting up the integral

The area between a curve, \( y = f(x) \), and an interval [a,b] on the x-axis can be found by calculating the definite integral: \( A = \int_{a}^{b} f(x) dx \). For our function, this becomes \( A = \int_{-1}^{1} (2-x^2) dx \).

Step 3: Calculate the integral

To calculate the integral, you need to find the anti-derivative of the function. The antiderivative of \(2 - x^2\) is \(2x - \frac{1}{3}x^3\). Apply the Fundamental Theorem of Calculus and calculate this antiderivative at 1 and -1, and subtract these values: \( A = [2-1^1.5/3] - [(2*(-1) - ((-1)^2)^1.5/3] = 2/3 \).

Step 4: Interpret the result

The area under the curve \(f(x) = 2 - x^2\) from x=-1 to x=1 is equal to \(2/3\) square units.

Key Concepts

Definite Integral

When we talk about finding the area under a curve on a graph, we are referring to a mathematical concept called the definite integral. This powerful tool is what we use to calculate the total 'accumulated' quantity, such as area, between a curve and the x-axis over a specific interval. Imagine slicing the area into infinitely thin vertical rectangles and then adding up their areas. The process of taking these infinitely small rectangles and finding the cumulative sum is essentially what makes up the definite integral.

The formula you'll often see is \( A = \int_{a}^{b} f(x) \, dx \), where \( A \) is the total area, \( a \), and \( b \) are the lower and upper bounds of the interval respectively, and \( f(x) \) is the function that defines the curve. In the example from the exercise, the function \( f(x) = 2 - x^2 \) is integrated between \( -1 \) and \( 1 \) to find the area under the parabola and above the x-axis.

Antiderivative

To solve a definite integral, we use the concept of an antiderivative, also known as the indefinite integral. This is the reverse process to differentiation, where we are seeking a function whose derivative is the given function \( f(x) \) from our integral. In other words, while differentiation gives us the slope of a function at any point, the antiderivative gives us the function that would, if differentiated, result in the original function.

An antiderivative is denoted by \( F(x) \) and is related to \( f(x) \) such that \( F'(x) = f(x) \). So, for our function \( 2 - x^2 \) the antiderivative is \( 2x - \frac{1}{3}x^3 \). We use the antiderivative to calculate the definite integral values at the bounds \( a \) and \( b \) of our interval, which allows us to find the area under the curve.

Fundamental Theorem of Calculus

The connection between the antiderivative and the definite integral is solidified by the Fundamental Theorem of Calculus. This pivotal theorem allows us to evaluate a definite integral using antiderivatives. It states that if a function \( f \) is continuous over an interval \( [a, b] \), and \( F \) is the antiderivative of \( f \) over that interval, then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). The theorem transforms the problem of finding the area under a curve into a much simpler task of just evaluating a function at two points.

In the exercise, by applying the Fundamental Theorem of Calculus to the antiderivative \( 2x - \frac{1}{3}x^3 \) between the bounds \( -1 \) and \( 1 \) we arrive at the result that the area is equal to \( \frac{2}{3} \) square units. This theorem is the link that integrates (no pun intended) antiderivatives with the practical application of calculating definite integrals.