Problem 61 - Calculus Early Transcendentals

Question

For each region \(R\), find the horizontal line \(y=k\) that divides \(R\) into two subregions of equal area. \(R\) is the region bounded by \(y=1-|x-1|\) and the \(x\) -axis.

Step-by-Step Solution

Answer

#tag_content# The value of k for the horizontal line y=k that divides the region R into two equal parts is found by following these steps: 1. Identify intersection points with the x-axis by setting y=0 and solving for x. 2. Calculate the total area under the curve using an integral. 3. Determine the x-value c that corresponds to half the area by solving an integral equation. 4. Find the k-value by substituting the x-value c back into the equation y=1-|x-1|. By following these steps, the value of k can be determined.

Step 1: Determine where the curve intersects the x-axis.

To find the intersection points, set \(y\) equal to \(0\) and solve for \(x\): $$0 = 1 - |x-1|$$ Solve for \(x\) when \(x-1 \ge 0\) and when \(x-1 < 0\) separately to obtain the intersection points. #Step 2: Calculate the total area under the curve#

Step 2: Set up and evaluate an integral to find the area under the curve.

The total area under the curve is given by the integral: $$A = \int_{a}^{b} (1 - |x-1|) dx$$ where \(a\) and \(b\) are the intersection points with the x-axis found in step 1. Calculate the area by splitting the integral into two parts based on the absolute value function. #Step 3: Determine the half-area line and the k-value#

Step 3: Find the line \(y=k\) that divides the region into two equal areas.

Since we want to find the horizontal line that divides the region into two equal areas, we can set the two equal areas as: $$\frac{A}{2} = \int_{a}^{c} (1-|x-1|) dx$$ where \(c\) is a value between the intersection point \(a\) and \(b\). Solve for the x-value \(c\) and substitute back into the equation \(y=1-|x-1|\) to find \(k\).