Problem 50 - Calculus Early Transcendentals

Question

Use the most efficient strategy for computing the area of the following regions. The region in the first quadrant bounded by \(y=x^{-1}, y=4 x,\) and \(y=x / 4\)

Step-by-Step Solution

Answer

Question: Determine the area of the region bounded by the curves \(y = x^{-1}, y = 4x,\) and \(y = x/4\). Answer: The area of the region bounded by the given curves is \(\frac{15}{8}\) square units.

Step 1: Finding the points of intersection

To find the points of intersection between the curves \(y = x^{-1}, y = 4x,\) and \(y = x/4\), we need to solve the following systems of equations: 1. \(x^{-1} = 4x\) 2. \(x^{-1} = x/4\) 3. \(4x = x/4\) Solving these equations gives us the coordinates of the points of intersection: 1. \((\frac{1}{2}, 2)\) 2. \((1, 1)\) 3. \((\frac{1}{4}, \frac{1}{4})\) Now that we have the points of intersection, we can set up the integral to compute the area.

Step 2: Setting up the integral for the area

To compute the area of the region, we can either integrate with respect to \(x\) or \(y\). Since the bounds of integration would be simpler with respect to \(y\), let's choose to integrate with respect to \(y\). The functions bounding the top and bottom of the region are \(y = 4x\) and \(y = x/4\). Solving each for \(x\) in terms of \(y\) gives us: 1. \(x = \frac{1}{4}y\) 2. \(x = 4y\) Since we are integrating with respect to \(y\), the bounds of the integration are determined by the minimum and maximum \(y\)-values of the points of intersection, which are \(\frac{1}{4}\) and \(2\). The area of the region can be computed using the integral: $$A = \int_{\frac{1}{4}}^{2} (0.25y - 4y) dy$$

Step 3: Evaluating the integral

To evaluate the integral, we need to compute the antiderivative of \((0.25y - 4y)\) with respect to \(y\): $$\int (0.25y - 4y) dy = (\frac{1}{8}y^2 - 2y^2) + C$$ Now, we can evaluate the definite integral: $$A = (\frac{1}{8}(2)^2 - 2(2)^2) - (\frac{1}{8}(\frac{1}{4})^2 - 2(\frac{1}{4})^2)$$ $$A = 2 - \frac{1}{8}$$ $$A = \frac{15}{8}$$ Thus, the area of the region bounded by the given curves is \(\frac{15}{8}\) square units.

Key Concepts

Intersection Points

When solving for the area between curves, the first essential step is determining where these curves intersect. The intersection points are crucial because they define the boundary of the area we are interested in. In our problem, we have three equations:

• \( y = x^{-1} \)
• \( y = 4x \)
• \( y = x/4 \)

To find the intersection points, we set pairs of these equations equal to each other. This method helps us find the exact points where the curves meet on the graph:

• Solve \( x^{-1} = 4x \) to find one intersection point.
• Solve \( x^{-1} = x/4 \) to find another intersection point.
• Solve \( 4x = x/4 \) to find the third intersection point.

Each of these equations will yield solutions for \( x \), and consequently, we can find the \( y \) values by substituting back into the original equations. This provides us with intersection points:

• \((\frac{1}{2}, 2)\)
• \((1, 1)\)
• \((\frac{1}{4}, \frac{1}{4})\)

These points are the corners of the region whose area we need to calculate.

Integration with respect to y

To find the area between the curves, integrating with respect to \( y \) can sometimes simplify the problem, especially when the curves change functions along the \( x \)-axis for which curve is on top. In this particular problem, integrating with respect to \( y \) is more straightforward because:

• The boundaries align more clearly with the transformation of functions.
• We have vertical lines in the first quadrant region bounded by solutions for \( x \).

To do this, we need to express \( x \) in terms of \( y \) for the bounding curves:

• From \( y = 4x \), solve to get \( x = \frac{1}{4}y \).
• From \( y = x/4 \), solve to get \( x = 4y \).

Now, the integral we set up will involve these expressions, with the lower bound \( y = \frac{1}{4} \) and the upper bound \( y = 2 \), the minimum and maximum \( y \)-values from the intersection points.

Definite Integral

The definite integral is the tool that allows us to compute the exact area between curves. When integrated over a specified interval, it sums the infinitesimally small areas under the curve, giving us the total area accurately. For our example, we have:

\[ A = \int_{\frac{1}{4}}^{2} \left( \frac{1}{4}y - 4y \right) \, dy \]
The difference inside the integral, \( (0.25y - 4y) \), represents the distance between the two functions after transforming them in terms of \( y \). Solving this integral involves finding its antiderivative:

• The antiderivative calculates as \( \frac{1}{8}y^2 - 2y^2 \).

By evaluating this from the lower to upper bounds, \( y = \frac{1}{4} \) to \( y = 2 \), we obtain the area. The result is calculated as:

• \( 2 - \frac{1}{8} \)
• \( \frac{15}{8} \) square units

First Quadrant Region

In calculus problems, specifying that a region is in the 'first quadrant' significantly simplifies analysis and setup. The first quadrant is where both \( x \) and \( y \) values are positive, which limits the functions and points we have to consider when finding areas between curves.

• This assumption allows us to disregard negative solutions for \( x \) or \( y \) that may arise when solving equation intersections.
• It ensures that all resulting areas calculations remain within the positive \( x \) and \( y \) plane, providing geometrically meaningful results.

In our exercise, all intersection points: \((\frac{1}{2}, 2)\), \((1, 1)\), and \((\frac{1}{4}, \frac{1}{4})\) fit within this quadrant. This allows us to confidently compute and evaluate using only the provided boundaries, ensuring the area computed is entirely within the desired region.