91Ó°ÊÓ

The inner integral is an essential part of evaluating a double integral. It is the integral we tackle first, and we choose an order of integration that simplifies our calculation. In this exercise, the inner integral focuses on integrating with respect to \( x \). Suppose you have the function \((2x + 3y)\) with the bounds from \( x = 0 \) to \( x = 2 \). Here, \( y \) is treated as a constant while solving the inner integral.

When integrating \( 2x + 3y \) with respect to \( x \), follow these steps:

• The integral of \( 2x \) with respect to \( x \) is \( x^2 \).
• The integral of \( 3y \) with respect to \( x \) is \( 3yx \), knowing \( y \) is a constant.

Calculate this from \( x = 0 \) to \( x = 2 \). Plug these values into your integrated function to find: \[[x^2 + 3yx]_{0}^{2} = (4 + 6y) - (0) = 4 + 6y\] This result, \( 4 + 6y \), becomes what you use for the next step.

After solving the inner integral, we move on to the outer integral. This integral is the second part of the double integral calculation and is now done with respect to \( y \). From the previous step, we found the expression \( 4 + 6y \), which is plugged into the outer integral. Applying the bounds from \( y = -1 \) to \( y = 1 \), integrate \( 4 + 6y \) with respect to \( y \):

• The integral of \( 4 \) with respect to \( y \) is \( 4y \).
• The integral of \( 6y \) with respect to \( y \) is \( 3y^2 \).

Evaluate this expression from \( y = -1 \) to \( y = 1 \), which gives: \[[4y + 3y^2]_{-1}^{1} = (4 + 3) - (-4 + 3) = 8\]The definite result of \( 8 \) is the value for the entire double integral.

Definite integrals are a crucial concept, representing the area under a curve within a specific interval. Unlike indefinite integrals, which include a constant of integration, definite integrals hold fixed upper and lower limits. For the given problem, each integration step is definite, specifying bounds that guide the calculation.

In our double integral \( \int_{-1}^{1} \int_{0}^{2} (2x + 3y) \; dx \; dy \), both the inner and outer integrals employ definite bounds:

• Inner integral: from \( x = 0 \) to \( x = 2 \)
• Outer integral: from \( y = -1 \) to \( y = 1 \)

These bounds indicate the region of interest for calculation.

Definite integrals not only provide a specific value (\( 8 \) in this exercise) but also help in understanding the concept of accumulation, portraying how quantities add up over defined sections of space or time.