Problem 42 - Applied Calculus

Question

For each function: a. Integrate ("by hand") to find the area under the curve between the given \(x\) -values. b. Verify your answer to part (a) by having your calculator graph the function and find the area (using a command like FnInt or \(\int \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\) ). $$ f(x)=9 x^{2}-6 x+1 \text { from } x=1 \text { to } x=2 $$

Step-by-Step Solution

Answer

The area under the curve is 13.

Step 1: Set Up the Integral

We are given the function \(f(x) = 9x^2 - 6x + 1\) and need to find the area under this curve from \(x = 1\) to \(x = 2\). The integral we need to evaluate is \(\int_{1}^{2} (9x^2 - 6x + 1) \, dx\).

Step 2: Find the Antiderivative

To integrate \(9x^2 - 6x + 1\), we find the antiderivative. The antiderivative of \(9x^2\) is \(3x^3\), the antiderivative of \(-6x\) is \(-3x^2\), and the antiderivative of \(1\) is \(x\). So, the antiderivative of \(9x^2 - 6x + 1\) is \(3x^3 - 3x^2 + x\).

Step 3: Evaluate the Definite Integral

Using the antiderivative \(3x^3 - 3x^2 + x\), we find the definite integral from \(x=1\) to \(x=2\). Compute \([3(2)^3 - 3(2)^2 + 2] - [3(1)^3 - 3(1)^2 + 1]\). Calculate: \([24 - 12 + 2] - [3 - 3 + 1]\). This gives \(14 - 1\), which equals 13.

Step 4: Verify Using a Calculator

Graph the function \(f(x) = 9x^2 - 6x + 1\) on a graphing calculator. Then, use a graphing calculator's function integration tool to find the area under the curve from \(x=1\) to \(x=2\). This should confirm our calculation as 13.

Key Concepts

Definite Integral

A definite integral is a powerful tool in calculus used for finding the area under a curve. Specifically, it represents the accumulation of quantities and is especially useful for calculating areas.

When we say 'finding the area under the curve,' this is exactly what a definite integral does. In our example exercise, the function given is \( f(x) = 9x^2 - 6x + 1 \) and we need to evaluate this from \( x = 1 \) to \( x = 2 \). This setup leads us to write the definite integral as:

• \( \int_{1}^{2} (9x^2 - 6x + 1) \, dx \)

The numbers at the bottom and top of the integral sign, known as the limits of integration, define the interval on the \( x \)-axis we are interested in.

After finding the antiderivative, we substitute the upper and lower limits to compute and evaluate the definite integral. This allows us to determine the precise area under the curve between these points, providing a numerical result of the bounded area.

Antiderivative

The antiderivative is the reverse process of differentiation. In other words, it's finding a function whose derivative gives the original function that you started with.

For calculating the definite integral, finding the antiderivative is a crucial step. Let's break down the process using the function \( f(x) = 9x^2 - 6x + 1 \) in our task:

• The antiderivative of \( 9x^2 \) is \( 3x^3 \).
• The antiderivative of \( -6x \) is \( -3x^2 \).
• The antiderivative of \( 1 \) is \( x \).

Putting these together, the antiderivative of \( 9x^2 - 6x + 1 \) is \( 3x^3 - 3x^2 + x \).

This function is necessary for evaluating the definite integral to find the area under a curve. Once the antiderivative is found, we substitute the bounds into it to find the specific numerical area.

Area Under the Curve

The area under the curve is a fundamental concept in calculus that refers to the region enclosed by the function, the \( x \)-axis, and the vertical lines determined by the limits of integration.

In our exercise, finding the area under \( f(x) = 9x^2 - 6x + 1 \) from \( x = 1 \) to \( x = 2 \) is done using integration. Here's how this unfolds:

• We calculate the antiderivative: \( 3x^3 - 3x^2 + x \).
• Evaluate it at the upper limit: \( x = 2 \).
• Evaluate it at the lower limit: \( x = 1 \).
• Subtract the lower from the upper: \( [3(2)^3 - 3(2)^2 + 2] - [3(1)^3 - 3(1)^2 + 1] \).

This results in an area calculation of \( 13 \), which means there are 13 square units under this specific segment of the curve.

By using both analytical and numerical methods, like graphing calculators, we can confirm this result. This process underscores the dependable nature of integration in measuring real-world quantities such as area.