91Ó°ÊÓ

A double integral is a way to integrate over a two-dimensional area. Imagine stacking an infinite number of infinitesimally thin sheets (or slices) over a plane. Double integrals give us the mathematical tool to sum up these sheets to find quantities like area, volume, or other aggregate measures associated with a function over a region. In our problem, we deal with the double integral of the function \[ f(x, y) = -x + 1 \] over the triangular region defined by its vertices. This is represented mathematically as: \[ \int_{y=0}^{y=2} \int_{x=0}^{x=y} (-x + 1) \, dx \, dy. \] This expression symbolizes collecting all the infinitely small contributions of the function over the specified region.

Calculating the area of a specific region is essential when determining average values of functions, as it forms part of the average value formula. In this exercise, the region of interest is a triangular area. The triangle, defined by vertices \((0,0), (0,2), (2,2)\), forms a right triangle with a base and height both equal to 2 units. The area of the triangle is determined using the formula for the area of a triangle:

• Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} \)

For our region, this becomes:\[ \frac{1}{2} \times 2 \times 2 = 2. \] This area value is used later to find the average value of the function over this region.

The process of function integration involves finding the integral of a function with respect to a variable. It is a fundamental operation in calculus used to calculate the total accumulation of a quantity. In the given problem, the function \(-x + 1\) is being integrated first with respect to \(x\), using limits from 0 to \(y\), which results in an integrated value of:\[ \left[ -\frac{x^2}{2} + x \right]_{0}^{y} = -\frac{y^2}{2} + y.\] This expression is then integrated with respect to \(y\), across its limits from 0 to 2, to finally produce:\[ \left[ -\frac{y^3}{6} + \frac{y^2}{2} \right]_{0}^{2} = \frac{4}{3}. \]Through these steps, the function's accumulation over the specified region is determined.

When integrating over a triangular region, understanding the limits and bounds of integration becomes important. Our triangle is defined in the \(xy\)-plane with the sides formed by \(y=0\), \(y=2\), and \(x=y\). The latter line, \(x=y\), sets the direction of the inner integral's upper limit. When setting up the double integral for this region, notice:

• For any fixed \(y\), \(x\) varies from 0 to \(y\).
• The integral bounds account for the shape of the triangle and ensure that every infinitesimal strip that accumulates contributes to the area between the slanted lines of the triangle.

This setup lets us utilize the boundaries of the triangle to precisely evaluate how the function \((-x + 1)\)averages out over the specified region.