Question

Use the most efficient strategy for computing the area of the following regions. The region in the first quadrant bounded by \(y=\frac{5}{2}-\frac{1}{x}\) and \(y=x\)

Short answer

Question: Find the area of the region bounded by \(y = \frac{5}{2} - \frac{1}{x}\) and \(y = x\) in the first quadrant. Answer: The area of the region bounded by the given curves in the first quadrant is 2 square units.

Step-by-step solution

Find the intersection point

To find the point of intersection, set both functions equal to each other: \(\frac{5}{2} - \frac{1}{x} = x\) Now, solve for \(x\): First, move all terms to one side: \(x^2 - \frac{5}{2} + \frac{1}{x} = 0\) Multiply both sides by \(2x\) to get rid of the fraction: \(2x^3 - 5x + 2 = 0\) Solving this cubic equation, we find that the point of intersection is: \((x, y) = \left(1, \frac{3}{2}\right)\)

Set up the integral

Now, we are ready to set up the integral to find the area between the two curves. We will use the formula: \(A = \int_a^b|f(x) - g(x)| dx\) Since the region is in the first quadrant and both functions are positive, we can simply write: \(A = \int_a^b (f(x) - g(x)) dx\) In this case, let \(f(x) = x\) and \(g(x) = \frac{5}{2} - \frac{1}{ x}\). Now, since we want the area of the region bounded by the curves, the limits of integration are the x-coordinates of the intersection point found in Step 1: 0 to 1.

Evaluate the integral

Plug in the functions and limits of integration to set up the integral and calculate the area: \(A = \int_0^1 (x - (\frac{5}{2} - \frac{1}{x})) dx\) Simplify the integral: \(A = \int_0^1 (x - \frac{5}{2} + \frac{1}{x}) dx\) Split the integral into separate parts: \(A = \int_0^1 x dx - \int_0^1 \frac{5}{2} dx + \int_0^1 \frac{1}{x} dx\) Now, find the antiderivatives and evaluate the integral: \(A = \left[\frac{1}{2}x^2\right]_0^1 - \left[\frac{5}{2}x\right]_0^1 + \left[\ln|x|\right]_0^1\) Substitute the limits of integration and find the difference: \(A = \left(\frac{1}{2}(1)^2 - \frac{1}{2}(0)^2\right) - \left(\frac{5}{2}(1) - \frac{5}{2}(0)\right) + \left(\ln(1) - \ln(0)\right)\) As ln(0) is undefined, we should take the limit as \(x\) approaches \(0\): \(A = \frac{1}{2} - \frac{5}{2} + \lim_{x\to 0} (\ln(1) - \ln(x))\) The last term in the equation evaluates to zero, as \(\ln(1) = 0\), so we have: \(A = \frac{1}{2} - \frac{5}{2}\)

Calculate the area

Finally, find the area by combining the terms: \(A = -2\) Since the area cannot be negative, take the absolute value of the result: \(A = 2\) Therefore, the area of the region bounded by the given curves in the first quadrant is 2 square units.

Key concepts

Integration

Integration is a fundamental concept in calculus that allows us to calculate areas under curves and solve many problems involving accumulation. It is essentially the reverse process of differentiation. For example, if we have a function that describes a curve on a graph, integration helps find the total area between this curve and the x-axis over a specified interval. This is done using the definite integral, which has limits of integration that define the region of interest.

In the given exercise, integration is used to find the area between two curves. The integral expression \( A = \int_a^b (f(x) - g(x)) \, dx \) is key here. This represents the accumulated area between two functions, \( f(x) \) and \( g(x) \), over the interval from \( a \) to \( b \). Subtraction inside the integral, \( f(x) - g(x) \), accounts for the actual difference in height between the two curves at each point along the x-axis, which is necessary to compute the area between them. By evaluating this integral, we obtain the desired area value, accounting for any possible overlaps or gaps between the curves.

Area between curves

The area between curves is a common application problem in calculus, where the goal is to determine the region enclosed by two or more curves. This helps solve real-world problems like finding the area of a field, or analyzing the space between geographical formations.

To find the area between two curves, you need to identify which curve is upper \( f(x) \), and which is lower \( g(x) \), within the interval of interest. The area is then calculated by integrating the difference \( (f(x) - g(x)) \) over the interval \([a, b]\), where the functions intersect. The problem from the exercise is a perfect example:

• Functions: \( y = x \) and \( y = \frac{5}{2} - \frac{1}{x} \)
• Intersection Points: Determine where the curves intersect to find \( x = 0 \) and \( x = 1 \).
• Integration: Integrate the difference from 0 to 1, \( \int_0^1 (x - (\frac{5}{2} - \frac{1}{x})) \, dx \).

This technique can be adapted to find areas between more complex curves or when dealing with more than two curves. It highlights the importance of setting correct limits and understanding the geometric relationship between the curves.

Cubic equation

Cubic equations are polynomial equations of the third degree, generally having the form \( ax^3 + bx^2 + cx + d = 0 \). Solving these equations is crucial in finding important points such as intersections which are vital in calculus problems involving curves.

In the exercise, finding the intersection of the curves leads to solving a cubic equation. Initially, both equations are set equal \( \frac{5}{2} - \frac{1}{x} = x \) to find the x-values where they meet. By manipulating this into a polynomial form, \( 2x^3 - 5x + 2 = 0 \), solving it reveals the intersection points necessary for defining our integration limits.

• The equation is transformed and solved to find values where the curves intersect.
• Identifies critical points for integrating the area between curves.
• The solution \((1, \frac{3}{2}) \) represents an exact point of intersection.

While this exercise results in a single real intersection in the relevant interval, cubic equations can have up to three real roots. Techniques like factoring, synthetic division, or the use of calculus may aid in solving them, especially when analytical solutions are not straightforward.