Problem 9 - Calculus Early Transcendentals

Question

Sketch the region and find its area. The region bounded by \(y=2(x+1), y=3(x+1),\) and \(x=4\)

Step-by-Step Solution

Answer

Question: Determine the area of the region enclosed by the equations \(y = 2(x+1)\), \(y = 3(x+1)\), and \(x=4\).

Step 1: Identify Intersection Points

First, we need to find the intersection points of the curves. These points are common for all three equations: 1. Intersection between \(y = 2(x+1)\) and \(y = 3(x + 1)\): Set \(2(x+1) = 3(x+1)\) and solve for \(x\). 2. Intersection between \(y = 2(x+1)\) and \(x = 4\): Substitute \(x = 4\) into the equation \(y = 2(x+1)\) to find the corresponding \(y\) value. 3. Intersection between \(y = 3(x+1)\) and \(x = 4\): Substitute \(x = 4\) into the equation \(y = 3(x+1)\) to find the corresponding \(y\) value.

Step 2: Sketch the Region

Using the intersection points we found in Step 1, we will now sketch the region enclosed by the three curves. 1. Plot the intersection points on the Cartesian plane. 2. Draw a line representing each of the given equations: \(y = 2(x+1)\), \(y = 3(x+1)\), and \(x = 4\). 3. Indicate the region enclosed by the equations.

Step 3: Set Up an Integral to Find the Area

We can find the area of the region enclosed by the curves by integrating the difference between the two linear functions from the intersection point on the left to the intersection point on the right: \(A = \int_{x_1}^{x_2} [(3(x + 1)) - (2(x+1))] dx \) Where \(x_1\) is the \(x\) value of the intersection point between \(y = 2(x+1)\) and \(y = 3(x+1)\), and \(x_2 = 4\) (the vertical line equation).

Step 4: Evaluate the Integral

Now we evaluate the integral and simplify: \(A = \int_{x_1}^{4} [(3x + 3) - (2x+2)] dx \) \(A = \int_{x_1}^{4} (x+1) dx \) By applying the power rule for integration, we get: \(A = \left[ \frac{x^2}{2} + x \right]_{x_1}^{4} \) Now, plug in the bounds \(x_1\) and \(4\) to obtain the area: \(A = \left[ \frac{(4)^2}{2} + 4 \right] - \left[ \frac{(x_1)^2}{2} + x_1 \right] \) Finally, substitute the value of \(x_1\) we calculated in Step 1 and evaluate the expression to find the area of the region.

Key Concepts

Intersection Points

The first step in solving the problem of finding the area between curves is determining the intersection points. These points mark the boundaries of the area we are examining. Here's how to identify them in this problem:

• Find where the lines intersect. For the functions \(y = 2(x+1)\) and \(y = 3(x+1)\), equate them: \(2(x+1) = 3(x+1)\), giving us an intersection point for \(x\).
• Calculate the intersection of each line with the vertical line \(x = 4\). For \(y = 2(x+1)\) and \(x = 4\), plug \(x = 4\) into the equation to find the \(y\) value.
• Repeat the process for \(y = 3(x+1)\) and \(x = 4\). Substitute \(x = 4\) into \(y = 3(x+1)\) to get the corresponding \(y\) value.

This process results in the specific intersection points that define the region we will analyze further. These points effectively set up the framework needed before moving on to graphing and integration.

Integral Setup

Once intersection points are identified, set up an integral to measure the area between the curves. This process involves calculating the vertical distance between the functions over the interval defined by these intersection points:

• First, determine the functions that form the upper and lower boundaries. The upper function, in this case, is \(y = 3(x+1)\), and the lower is \(y = 2(x+1)\).
• The integral is defined as the integral of the difference between these functions from the leftmost intersection point to the vertical line \(x=4\): \[ A = \int_{x_1}^{4} [(3(x + 1)) - (2(x+1))] \, dx \]
• This expression captures the area by summing small infinitesimally thin slices vertically between the functions over the defined interval.

By understanding integral setup, you are explicitly defining the area you intend to measure. It's crucial because it indicates the exact part of the function space that corresponds to the region enclosed by the curves.

Evaluating Integrals

After setting up the integral, evaluate it to find the specific area. Evaluating integrals is essentially summing up these vertical slices over the defined range to get the total area. Follow these steps:

• Start by simplifying the expression within the integral: \[ A = \int_{x_1}^{4} (x+1) \, dx \]
• The integration of \(x+1\) involves applying basic integration rules: \[ A = \left[ \frac{x^2}{2} + x \right]_{x_1}^{4} \]
• Substitute the boundary values of \(x\), starting from the intersection point \(x_1\) and ending at \(x = 4\). This results in the formula: \[ A = \left[ \frac{4^2}{2} + 4 \right] - \left[ \frac{x_1^2}{2} + x_1 \right] \]

The final arithmetic will give you the specific area enclosed between the curves. The integration here involves understanding basic calculus techniques, capturing the cumulative nature of the operation. Ultimately, it's the numerical difference that signifies the area between the two curves.