91Ó°ÊÓ

When dealing with double integrals, one key concept is understanding the region over which you're integrating. This is essential because it determines the limits of integration and helps us visualize what we are calculating. In this problem, we look at \( \int _ { - 1 } ^ { 2 } \int _ { y ^ { 2 } } ^ { y + 2 } d x d y \), which defines a region in the \( xy \)-plane. This means our interest lies in the two-dimensional space bounded by these limits. The integral describes iterating over \( y \) from \(-1 \) to \( 2 \), defining the vertical span, and for each specific value of \( y \), \( x \) values range from \( y^2 \) to \( y + 2 \). Knowing this, you can visualize the domain as a vertical strip that has a pre-defined bottom edge, top edge, left boundary, and right boundary.

In this exercise, finding where the two curves intersect is crucial. It allows us to determine the exact region enclosed by these curves. We look at the equations of the bounding curves:

• \( x = y^2 \), a parabola opening to the right,
• \( x = y + 2 \), a straight line.

To find their points of intersection, we set the equations equal, \( y^2 = y + 2 \), and solve for \( y \). Simplifying gives a quadratic \( y^2 - y - 2 = 0 \). Upon factoring, we get \( (y - 2)(y + 1) = 0 \), leading to solutions \( y = 2 \) and \( y = -1 \). These values substitute back into either equation to find corresponding \( x \)-values, giving points \((4, 2)\) and \((1, -1)\) as the intersections. These points help us neatly define the region of interest.

Calculating the area of the region enclosed by the curves is the goal of our integral. Once we identify the region in the \( xy \)-plane and the bounds for \( y \) and \( x \), we can set up our double integral to find the area. Our task here includes:1. Evaluating the iterated integral in steps.2. First, perform the inner integral: \( \int _{ y^2 } ^{ y+2 } 1 \, dx \), which simplifies to \((y+2 - y^2)\).3. Then, integrate this result over \( y \) from \(-1\) to \( 2 \): \[ \int _{-1}^{2} (y+2 - y^2) \, dy \].Breaking it down, integrate each part separately to get the sum \( \frac{9}{2} \) square units as the total area. It involves some basic integration rules to compute these efficiently, ensuring every student can follow along and replicate at this point.

Bounding curves are essential in analyzing and solving integration problems that deal with regions in the \( xy \)-plane. For this exercise, our bounding curves are \( x = y^2 \) and \( x = y + 2 \). Understanding how each boundary functions helps to accurately describe the area we wish to calculate.

• The curve \( x = y^2 \) forms a parabola opening to the right and acts as the left boundary in our region.
• The curve \( x = y + 2 \) is a straight line with a positive linear slope, serving as the right boundary.

Between \( y = -1 \) and \( y = 2 \), the region of interest stretches horizontally between these curves. By setting up and solving the problems with these bounding curves in mind, students can see their role in determining the region over which we perform the integration.