Problem 105 - University Calculus: Early Transcendentals

Question

Find the area of the region in the first quadrant bounded on the left by the \(y\) -axis, below by the line \(y=x / 4,\) above left by the curve \(y=1+\sqrt{x},\) and above right by the curve \(y=2 / \sqrt{x}\).

Step-by-Step Solution

Answer

Calculate the integrals from 0 to 4 and 4 to 12 + 8√2 for the given bounds. Sum them to get the total area.

Step 1: Determine the Points of Intersection

First, find the intersection points between the lines and curves involved.1. **For** \(y=\frac{x}{4}\) and \(y=1+\sqrt{x}\): Equate: \(\frac{x}{4} = 1 + \sqrt{x}\) Rearrange to solve: \[x = 4 + 4\sqrt{x}\] Let \(z = \sqrt{x}\), then \(x = z^2\): \[z^2 - 4z - 4 = 0\] Solve this quadratic: \[z = \frac{4 \pm \sqrt{16 + 16}}{2} = \frac{4 \pm \sqrt{32}}{2}\] Simplifies to: \[z = 2 \pm 2\sqrt{2}\] Only the positive root is valid since the first quadrant only has positive values: \[z = 2 + 2\sqrt{2}\] Thus, \(x = (2 + 2\sqrt{2})^2 = 4 + 8\sqrt{2} + 8 = 12 + 8\sqrt{2}\).2. **For** \(y=\frac{x}{4}\) and \(y=\frac{2}{\sqrt{x}}\): Equate: \(\frac{x}{4} = \frac{2}{\sqrt{x}}\) Solve: \[x^{3/2} = 8\] Taking cube roots: \[x = 4\].Thus, key intersection points are \((12 + 8\sqrt{2}, 1+2\sqrt{2})\) between the first set and \((4, 1)\) for the second set.

Step 2: Set Up the Integral Boundaries

Identify the region boundaries in the first quadrant:- **Left Boundary**: \(y = \frac{x}{4}\)- **Below Boundary**: The curve starts at the y-axis; hence, \(y = \frac{x}{4}\).- **Upper Left Boundary**: \(y=1+\sqrt{x}\)- **Upper Right Boundary**: \(y=\frac{2}{\sqrt{x}}\)The integral limits will be separated at \(x=4\) and \(x=12 + 8\sqrt{2}\) based on the intersection points.

Step 3: Integrate to Find the Area (0 to 4)

Integrate the difference between the upper right and lower boundary from \(x=0\) to \(x=4\):\[\int_0^4 \left(\frac{2}{\sqrt{x}} - \frac{x}{4}\right)\,dx\]1. Solve \(\int \frac{2}{\sqrt{x}}\,dx\): Which is: \(\int 2x^{-1/2} \, dx = 4x^{1/2}\).2. Solve \(\int \frac{x}{4}\,dx\): Which is: \(\frac{1}{8}x^2\).Evaluate: \[4x^{1/2} - \frac{1}{8}x^2\Bigg|_0^4 = (8 - 2) - (0 - 0) = 6\].

Step 4: Integrate to Find the Area (4 to 12+8√2)

Integrate the difference between the upper left and lower boundary from \(x=4\) to \(x=12 + 8\sqrt{2}\):\[\int_4^{12 + 8\sqrt{2}} \left(1+\sqrt{x} - \frac{x}{4}\right)\,dx\]1. Solve \(\int (1+\sqrt{x}) \, dx\): Which is: \(x + \frac{2}{3}x^{3/2}\).2. Solve \(\int \frac{x}{4}\,dx\): Which is: \(\frac{1}{8}x^2\).Evaluate: \[(12 + 8\sqrt{2} + \frac{2}{3}(12 + 8\sqrt{2})^{3/2} - \frac{1}{8}(12 + 8\sqrt{2})^2)\]Subtract the area contribution at 4 found in previous steps.

Step 5: Calculate the Integral Results

Now calculate the value of each integral- For the first integral result: \(6\).- For the second integral, input the value found into the formulas:Substitute and simplify to find an approximated value **Final Area**:The sum of both integrated sections results

Key Concepts

Area of a Region

Calculating the area of a region bounded by different curves and lines is a common task in calculus. Understanding this helps us determine how much space lies within specific boundaries on a graph.
To find the area of a region, we typically need to:

• Identify the boundaries: These can be lines, curves, or both, as seen in the example where the region is bound by a line, two curves, and the y-axis.
• Determine the points of intersection: These points are crucial because they often define the limits of our integration. In this exercise, intersections occur where lines meet the curves.

By understanding these intersections, we can better visualize the sectioned-off areas. The goal is to set up integrals that precisely calculate the area within these boundaries.
If we correctly identify the intersections and insert correct limits into our integrals, we solve accurately--just like in the problem which calculated the area first from 0 to 4, then 4 to our calculated intersection at approximately 12 + 8\(\sqrt{2}\).

Integration

Integration is a method used in calculus to determine the accumulated quantity, like area or volume, under a curve. It essentially adds up infinitely small pieces to give us a big picture. In this exercise:

• We use definite integrals to find precise areas between curves and lines.
• Each integral corresponds to a section bounded by intersections.
• The integrals were split into two parts: one from 0 to 4 and the other from 4 to the intersection point at about 12 + 8\(\sqrt{2}\).

The formula for integration involves evaluating the integral at its limits, often using antiderivatives. For instance, integrating \(\int \frac{2}{\sqrt{x}}\,dx\) gives \(4x^{1/2}\), because this function, when derived, returns the original function.
This process continues for each curve and boundary, calculated separately, before summing up the areas to find the final bounded region's area. Integration is powerful because it allows us to tackle non-linear boundaries easily.

Quadratic Equations

Quadratic equations, usually in the form \(ax^2 + bx + c = 0\), appear frequently in calculus and are critical in solving for exact points of intersection in many problems.
In this exercise, the intersection between two curves and a line required solving a quadratic equation:

• By letting \(z=\sqrt{x}\), the intersection problem \(\frac{x}{4} = 1 + \sqrt{x}\) simplified into a quadratic equation: \(z^2 - 4z - 4 = 0\).
• Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we found the roots, keeping in mind we're only interested in positive roots for intersections in the first quadrant.

Solving this helped identify the boundaries correctly, crucial for planning the integration. Understanding quadratic equations ensures we accurately find and use these intersections, providing clarity on where boundaries change or intersect.