Express the area of the shaded region in Exercise 5 as the sum of two integrals with respect to y. Do not evaluate the integrals.

Verified step by step guidance

1. Identify the region bounded by the lines \(y = 2 - x\) and \(y = x\), and the vertical line \(x = 1\). The vertices of the shaded triangular region are at points \((0, 2)\), \((1, 1)\), and \((1, 0)\).
2. Since the problem asks for the area expressed as the sum of two integrals with respect to \(y\), we need to rewrite the boundary curves in terms of \(x\) as functions of \(y\). From \(y = 2 - x\), solve for \(x\): \(x = 2 - y\). From \(y = x\), solve for \(x\): \(x = y\).
3. Determine the range of \(y\) values for the two parts of the region. The lower part of the region extends from \(y = 0\) to \(y = 1\), and the upper part extends from \(y = 1\) to \(y = 2\).
4. For \$0 \leq y \leq 1\(, the left boundary is \)x = y\( and the right boundary is \)x = 1\(. So, the area contribution here is the integral \)\int_0^1 (1 - y) \, dy$.
5. For \$1 \leq y \leq 2\(, the left boundary is \)x = 0\( and the right boundary is \)x = 2 - y\(. So, the area contribution here is the integral \)\int_1^2 (2 - y) \, dy$. The total area is the sum of these two integrals.

Key Concepts

Setting up integrals with respect to y

When expressing area as integrals with respect to y, the region is sliced horizontally. This requires rewriting the boundary curves as functions of y (i.e., x in terms of y) and integrating over the y-intervals that cover the region. This approach is useful when the region is bounded more naturally by horizontal slices.

Finding inverse functions of boundary curves

To integrate with respect to y, the given functions y = 2 - x and y = x must be inverted to express x as functions of y. For example, from y = 2 - x, we get x = 2 - y, and from y = x, we get x = y. These inverse functions define the horizontal boundaries of the region for integration.

Partitioning the region for multiple integrals

The shaded region is divided into two parts along y = 1 because the left and right boundaries change at this point. Each part corresponds to a different pair of boundary functions in terms of y, so the total area is expressed as the sum of two integrals over different y-intervals.