Problem 18 - Precalculus: Mathematical for Calculus

Question

Find the area of the region that lies under the graph of \(f\) over the given interval. $$f(x)=20-2 x^{2}, \quad 2 \leq x \leq 3$$

Step-by-Step Solution

Answer

The area under the graph from 2 to 3 is approximately 7.333 square units.

Step 1: Set up the integral

To find the area under the curve of the function \( f(x) = 20 - 2x^2 \) from \( x = 2 \) to \( x = 3 \), we first need to set up the integral: \[\int_{2}^{3} (20 - 2x^2) \, dx\]

Step 2: Find the antiderivative

Determine the antiderivative of the function \( 20 - 2x^2 \). The antiderivative of \( 20 \) is \( 20x \), and the antiderivative of \( -2x^2 \) is \( -\frac{2}{3}x^3 \). Thus, the antiderivative is: \[ F(x) = 20x - \frac{2}{3}x^3 \]

Step 3: Evaluate the antiderivative at the upper limit

Substitute \( x = 3 \) into the antiderivative: \[ F(3) = 20(3) - \frac{2}{3}(3)^3 = 60 - \frac{54}{3} = 60 - 18 = 42 \]

Step 4: Evaluate the antiderivative at the lower limit

Substitute \( x = 2 \) into the antiderivative: \[ F(2) = 20(2) - \frac{2}{3}(2)^3 = 40 - \frac{16}{3} = 40 - 5.333 = 34.667 \]

Step 5: Calculate the area

Compute the area under the curve by subtracting the value of the antiderivative at the lower limit from the value at the upper limit:\[ \text{Area} = F(3) - F(2) = 42 - 34.667 = 7.333 \]

Key Concepts

Definite Integral

A definite integral is a core concept in calculus that allows us to calculate the area under a curve within specific limits. In simpler terms, it helps us find the total "sum" of infinitely small slices of area from the starting point to the end point along the curve of a function.
When we talk about the area under a curve for a function like \( f(x) = 20 - 2x^2 \) over the interval from \( x = 2 \) to \( x = 3 \), we use the definite integral to compute this region. The integral is set up as follows:

• The definite integral has an upper and lower limit, in this case, 3 and 2.
• The notation is \( \int_{2}^{3} (20 - 2x^2) \, dx \).

After calculating the definite integral, we determine the exact area enclosed, which is often crucial in various applications, from physics to economics.

Antiderivative

In calculus, an antiderivative of a function is essentially the reverse of taking a derivative. It's a function whose derivative equals the original function. When solving problems related to definite integrals, finding the antiderivative is a crucial step.
To find the area under the curve, we first determine the antiderivative of the given function \( f(x) = 20 - 2x^2 \). The antiderivative, denoted by \( F(x) \), takes the form:

• For the constant \( 20 \), the antiderivative is \( 20x \).
• For \( -2x^2 \), the antiderivative is \( -\frac{2}{3}x^3 \).

Combining these results, we get the antiderivative \( F(x) = 20x - \frac{2}{3}x^3 \). This antiderivative will later be evaluated at the upper and lower limits to compute the area.

Integration in Calculus

Integration is a fundamental concept in calculus that allows us to "add up" an infinite number of small slices to find areas, among other things. It is often described as the inverse operation to differentiation. There are several motivations and methods for integration:

• Purpose: To find areas, volumes, central points, and many useful quantities.
• Methods: Indefinite integration (finding an antiderivative) and definite integration (using limits to find an exact area).
• Applications: Widely applied in physics, engineering, and economics.

In our problem, integration involves both finding the antiderivative (indefinite integration) and applying limits (definite integration). Using these steps, we can determine the exact area under the function from \( x = 2 \) to \( x = 3 \), providing a simple technique to solve complex real-world problems.