14–25. {Use of Tech} Areas of regions Determine the area of the given region.


The region in the first quadrant bounded by y = x/6 and y = 1−|x/2−1|

Verified step by step guidance

1. First, identify the curves that bound the region in the first quadrant: the line \(y = \frac{x}{6}\) and the piecewise function \(y = 1 - \left| \frac{x}{2} - 1 \right|\).
2. Next, analyze the function \(y = 1 - \left| \frac{x}{2} - 1 \right|\) by considering the expression inside the absolute value. Set \(\frac{x}{2} - 1 = 0\) to find the critical point \(x = 2\) where the function changes its form.
3. Rewrite \(y = 1 - \left| \frac{x}{2} - 1 \right|\) as a piecewise function: For \(x \leq 2\), \(y = 1 - (1 - \frac{x}{2}) = \frac{x}{2}\), For \(x > 2\), \(y = 1 - (\frac{x}{2} - 1) = 2 - \frac{x}{2}\).
4. Determine the points of intersection between \(y = \frac{x}{6}\) and each piece of the piecewise function: - For \(x \leq 2\), solve \(\frac{x}{6} = \frac{x}{2}\), - For \(x > 2\), solve \(\frac{x}{6} = 2 - \frac{x}{2}\). These intersection points will serve as limits of integration.
5. Set up the integral(s) for the area of the region in the first quadrant bounded by the curves. The area can be found by integrating the difference between the upper and lower functions over the appropriate intervals determined by the intersection points.

Key Concepts

Understanding Piecewise and Absolute Value Functions

The function y = 1 − |x/2 − 1| involves an absolute value, which creates a piecewise definition. Understanding how to rewrite and interpret absolute value expressions as piecewise linear functions is essential to identify the shape and boundaries of the region.

Finding Points of Intersection

Determining the area bounded by two curves requires finding their points of intersection. Solving y = x/6 and y = 1 − |x/2 − 1| simultaneously helps establish the limits of integration for calculating the area.

Definite Integration to Find Area Between Curves

The area between two curves is found by integrating the difference of the upper and lower functions over the interval defined by their intersection points. Setting up and evaluating the definite integral accurately yields the desired area.