Problem 26 - Applied Calculus

Question

Use a definite integral to find the area under each curve between the given \(x\) -values. For Exercises \(19-24\) also make a sketch of the curve showing the region. $$ f(x)=6 x^{2} \text { from } x=2 \text { to } x=3 $$

Step-by-Step Solution

Answer

The area under the curve is 38 units squared.

Step 1: Understand the Problem

We need to find the area under the curve of the function \( f(x) = 6x^2 \) from \( x = 2 \) to \( x = 3 \). This requires evaluating a definite integral to find the area of the region under the curve between these two points.

Step 2: Set Up the Definite Integral

The area under the curve is given by the definite integral of the function over the interval from \( x=2 \) to \( x=3 \). We set this up as follows: \[ \int_{2}^{3} 6x^2 \, dx \]

Step 3: Find the Antiderivative

To solve the integral, we first find the antiderivative of \( f(x) = 6x^2 \). The antiderivative of \( 6x^2 \) is \( 2x^3 \) because the power rule for integration tells us to increase the exponent by 1 and then divide by the new exponent.

Step 4: Evaluate the Definite Integral

The definite integral from \( x=2 \) to \( x=3 \) of \( 6x^2 \) is calculated by evaluating the antiderivative at the upper and lower bounds and subtracting:\[ \int_{2}^{3} 6x^2 \, dx = \left[ 2x^3 \right]_2^3 = 2(3)^3 - 2(2)^3 \].

Step 5: Complete the Calculation

First, compute \( 2(3)^3 = 54 \) and \( 2(2)^3 = 16 \). Then, subtract the lower bound evaluation from the upper bound evaluation: 54 - 16 = 38.

Key Concepts

Antiderivative

The antiderivative is an essential concept when solving integrals. It is sometimes called the indefinite integral. The process involves finding a function that, when differentiated, gives the original function. In this exercise, we need the antiderivative of the function \( f(x) = 6x^2 \). This can be done using the power rule, which states that if \( f(x) = x^n \), then its antiderivative is \( \frac{x^{n+1}}{n+1} \).

Therefore, for \( 6x^2 \), the antiderivative becomes \( 2x^3 \). This happens because we increase the power of \( x \) from 2 to 3 and then divide the coefficient 6 by the new power 3. It's important to understand that finding an antiderivative is a way to reverse the process of differentiation.

• Antiderivative is a type of integration
• Used to reverse differentiation
• Key for solving definite integrals

Area Under a Curve

Finding the area under a curve is a common use of definite integrals. This area tells us how much space lies beneath the curve of a function within a certain interval on the \( x \)-axis.

For example, in our exercise, we are finding the area under \( f(x) = 6x^2 \) from \( x=2 \) to \( x=3 \). To determine this area, we use the definite integral \( \int_{2}^{3} 6x^2 \, dx \). Calculating this defined area allows us to measure how much content resides under the curve over a specified interval.

• Visualizes the integral concept
• Represents exact area calculation
• Essential for real-world applications

Integration

Integration is a fundamental concept in calculus. It is the process of finding integrals, which can be viewed as the opposite operation of differentiation. Integration can be indefinite or definite. Indefinite integrals are related to finding antiderivatives, and definite integrals allow for the computation of the area under a curve.

In our example, by integrating \( 6x^2 \) from \( x=2 \) to \( x=3 \), we found a definite integral. The steps involved:

• Finding the antiderivative \( 2x^3 \)
• Evaluating this antiderivative at the bounds of \([2, 3]\)
• Subtraction to find the area (\(54 - 16 = 38\))

Integration binds the concepts of antiderivatives and area together, providing powerful tools for analyzing and understanding functions.