Problem 26 - Applied Calculus

Question

Find the area between the graph of \(y=x^{2}-2\) and the \(x\) -axis, between \(x=0\) and \(x=3\).

Step-by-Step Solution

Answer

The area is 3 square units.

Step 1: Understand the Problem

We need to calculate the area between the curve represented by the function \(y = x^2 - 2\) and the \(x\)-axis, specifically from \(x = 0\) to \(x = 3\). The area can be found by integrating the function over the given interval.

Step 2: Set Up the Integral

The area under the curve and above the \(x\)-axis can be calculated using the integral:\[ \int_{a}^{b} (x^2 - 2) \, dx \]where \(a = 0\) and \(b = 3\).

Step 3: Calculate the Definite Integral

We solve the integral to find the area:\[ \int_{0}^{3} (x^2 - 2) \, dx = \left[ \frac{x^3}{3} - 2x \right]_{0}^{3} \]First, we substitute \(x = 3\) into the antiderivative:\[ \left( \frac{3^3}{3} - 2 \times 3 \right) = \left( 9 - 6 \right) = 3 \]Next, substitute \(x = 0\):\[ \left( \frac{0^3}{3} - 2 \times 0 \right) = 0 \]

Step 4: Compute the Area

Subtract the results obtained when substituting the limits:\[ 3 - 0 = 3 \]Thus, the area between the curve and the \(x\)-axis from \(x = 0\) to \(x = 3\) is \(3\) square units.

Key Concepts

Area Under a Curve

When dealing with the concept of the area under a curve, we are essentially looking at a visual representation of the space between a curve and a specific part of the x-axis. In mathematical terms, this can be understood as the definite integral of a function over a particular interval. But don't worry, that’s just a fancy way to explain a process that isn't so scary. To break it down, imagine plotting the function on a graph. The area under this curve is like coloring in all the space beneath the line and above the x-axis. For this to make sense, we need two boundaries on the x-axis, which we call the limits of integration. In our example, these are from x=0 to x=3. Finding this area is crucial in several applications like physics for determining distance or economics for calculating profit from certain models. So, just remember: the area under a curve is all about figuring out how much space exists between the curve and the x-axis over a set interval. The secret to unlocking this lies in integration.

Antiderivative

The antiderivative is closely related to the concept of the derivative, but it works in reverse. When we talk about finding the antiderivative of a function, we're searching for another function whose derivative gives us the original function. This process is often called 'integration'. Why is the antiderivative important in calculating areas? Because when we find the definite integral of a function, we are essentially seeking the antiderivative of that function to help calculate the total area under the curve. Consider the function in our exercise, y = x² - 2. To find its antiderivative, we add one to the exponent of each term and divide by this new exponent for polynomials. Therefore, the antiderivative of x² is x³/3, and for the -2, it becomes -2x. Applying these rules allows us to evaluate the definite integral by computing the antiderivative function values at specified limits (here, x=0 and x=3). Recognizing the role of the antiderivative helps in understanding how the integration process unveils the area under the curve.

Integration

Integration is the key mathematical tool used to find the area under a curve over a specific interval. It acts like the opposite of differentiation. While differentiation slices functions into smaller parts, integration essentially puts those pieces back together to find the whole. When we integrate a function like y = x² - 2 from x = 0 to x = 3, we are determining the sum of infinitesimally small areas that lie between the curve and the x-axis. This operation results in a numerical value that represents the total area. The process involves setting up a definite integral, which is expressed as the integral of our function between two limits. This involves two steps:

• Finding the antiderivative (or indefinite integral) of the function.
• Using the Fundamental Theorem of Calculus to calculate the difference of the antiderivative evaluated at these upper and lower limits.

So, as seen in our example, when using integration, once we find the antiderivative, we plug in the values to find the precise area under the curve. Integration is an elegant and powerful way to solve real-world problems involving areas and accumulations.