91Ó°ÊÓ

Double integrals are a fundamental concept in multivariable calculus. They allow us to compute the volume under a surface over a given region. Instead of integrating a function of one variable, we integrate a function of two variables over a two-dimensional region.

In our problem, the double integral is written as:
\( \int_{-2}^{0}\left( \int_{-1}^{1} x e^{x y} d y\right) d x \)

Here, we are integrating first with respect to \( y \) and then with respect to \( x \). This approach is called iterated integration. The general form for a double integral is:
\( \iint\_D f(x, y) dA \)

where \( D \) is the region of integration and \( dA \) represents the differential area element.
Double integrals can be evaluated in two main methods:

• Iterated integrals, where you integrate one variable at a time.
• Changing the order of integration, which may simplify the calculation.

Our given problem follows the iterated integrals approach, which we shall discuss in the integration and exponential function sections.

Integration is a core concept in calculus, used to find the accumulated quantity of a function over a region. In our problem, we deal with integrating an exponential function.

First, we integrate the inner integral:

\( \int_{-1}^{1} x e^{x y} d y \).
Here, \( x \) is treated as a constant, and we need to compute the integral of \( e^{x y} \) with respect to \( y \).

The integral of \( e^{x y} \) with respect to \( y \) is:

\( \frac{e^{x y}}{x} \).

This results in:

\( x \left \frac{e^{x y}}{x} \right |_{-1}^{1} \ = e^{x} - e^{-x} \).

Next, we handle the outer integral:

\( \int_{-2}^{0} (e^{x} - e^{-x}) d x \),

by breaking it into two simpler parts:
\( \int_{-2}^{0} e^{x} d x - \int_{-2}^{0} e^{-x} d x \).
Calculating each part:

\( \int_{-2}^{0} e^{x} d x = e^{0} - e^{-2} = 1 - \frac{1}{e^{2}} \)
\( \int_{-2}^{0} e^{-x} d x = -e^{0} + e^{2} = e^{2} - 1 \)

Combining these results, we get:

\( 2 - (e^{2} + \frac{1}{e^{2}}) \).

The exponential function, denoted as \( e^{x} \), is a crucial function in mathematics that models continuous growth or decay. Its key properties include:

• The function is always positive.
• It has a constant rate of growth proportional to its value.
• The derivative and integral of \( e^{x} \) are both \( e^{x} \).

In our integral problem, \( e^{x y} \) is used. Here's how it works:

• When integrating \( e^{x y} \) with respect to \( y \), holding \( x \) constant, we utilize
  the rule:
  \( \int e^{k y} d y = \frac{e^{k y}}{k} \) where \( k \) is a constant.

Applying this to our inner integral:
\( \int_{-1}^{1} x e^{x y} d y \),
we find that\( \int_{-1}^{1} x e^{x y} d y \) evaluates to:
\( e^{x} - e^{-x} \).

When integrating \( e^{x} \) and \( e^{-x} \), we use:

\( \int e^{k x} d x = \frac{e^{k x}}{k} \), leading to solutions:
\( 1 - \frac{1}{e^{2}} \) and
\( e^{2} - 1 \).

By understanding these key points about exponential functions, we can effectively tackle integrals involving such terms.