Question

Evaluate $$ \int_{0}^{1} \int_{0}^{1} x y e^{x^{2}+y^{2}} d y d x $$

Short answer

The integral evaluates to \( \frac{(e-1)^2}{4} \).

Step-by-step solution

Identify the Order of Integration

The order of integration is from inside to outside: the inner integral with respect to y and the outer integral with respect to x. We'll start by evaluating the inner integral wrt y.

Evaluate the Inner Integral

Consider the integral \( \int_{0}^{1} y e^{x^{2}+y^{2}} \, dy \). Use substitution: let \( u = x^2 + y^2 \), then \( du = 2y \, dy \) or \( y \, dy = \frac{1}{2} \, du \). The limits change from \( y = 0 \rightarrow u = x^2 \) to \( y = 1 \rightarrow u = x^2 + 1 \). Integrate to get \( \frac{1}{2} \int_{x^2}^{x^2+1} e^u \, du = \frac{1}{2} [e^{x^2+1} - e^{x^2}] \).

Evaluate the Outer Integral

The integral now becomes \( \int_{0}^{1} x \frac{1}{2} [e^{x^2+1} - e^{x^2}] \, dx \). This can be split into two separate integrals: \( \frac{1}{2} \left( e \int_{0}^{1} x \, e^{x^2} \, dx - \int_{0}^{1} x \, e^{x^2} \, dx \right) \).

Integrate e^{x^2} Terms

For \( \int x \, e^{x^2} \, dx \), use substitution: let \( v = x^2 \), then \( dv = 2x \, dx \), or \( x \, dx = \frac{1}{2} \, dv \). The integral becomes \( \frac{1}{2} \int e^v \, dv = \frac{1}{2} e^v + C \). Back substitute \( v = x^2 \) to get \( \frac{1}{2} e^{x^2} \). Evaluate from 0 to 1 gives \( \frac{1}{2} [e - 1] \).

Substitute Back to Solve

Plug the results back into the outer integral: \( \frac{1}{2} \left( e \cdot \frac{1}{2} [e - 1] - \frac{1}{2} [e - 1] \right) = \frac{1}{4} [e(e - 1) - (e - 1)] \). Simplify this to \( \frac{1}{4} [e^2 - e - e + 1] = \frac{1}{4} [e^2 - 2e + 1] = \frac{1}{4} (e - 1)^2 \).

Final Simplification

The integral result is \( \frac{(e-1)^2}{4} \). This expression represents the area under the surface described by the given double integral.

Key concepts

Substitution Method

The substitution method is a powerful technique for evaluating integrals, especially when dealing with composite functions. In the original exercise, the substitution method simplifies both the inner and outer integrals by transforming them into a more manageable form.
When substituting, we replace a portion of the integral with a new variable to simplify the expression. For instance, in the given exercise, for the inner integral, the substitution \( u = x^2 + y^2 \) was made. This helps in reducing complexity since \( e^{u} \) is simpler to integrate than \( e^{x^2 + y^2} \).
It's important to adjust the limits of integration to reflect the new variable. So, when \( u = x^2 + y^2 \), the limits change accordingly. Substitution often involves these key steps:

• Choosing a substitution \( u = g(y) \) or \( u = g(x) \)
• Finding \( du \), where \( du = g'(x) \, dx \)
• Adjusting the integration limits
• Solving the simpler integral \( \int f(u) \, du \)

In our context, using substitution transformed a cumbersome integral into a straightforward calculation.

Iterated Integrals

Iterated integrals involve evaluating a double integral by integrating one variable at a time. This method is crucial for solving problems involving areas or volumes under surfaces described by a function of two variables.
In the exercise above, the iterated integral method is applied by first integrating with respect to \( y \) and then \( x \). These are also known as inner and outer integrals, respectively. The process follows a structured approach:

• First, solve the inner integral with respect to \( y \), treating \( x \) as a constant.
• Next, use the result to solve the outer integral with respect to \( x \).

It is worth noting that the order of integration (i.e., \( dy \, dx \)) is given by the limits of integration provided in the original problem, often requiring thoughtful consideration to ensure the limits correspond properly to the function being integrated.

Integration Techniques

Integration techniques refer to the various methods used to solve integral problems, particularly when dealing with complex functions. In the given exercise, multiple techniques are combined to simplify the problem.
The combination of substitution and integration by parts often help make challenging integrals solvable:

• Substitution allows simplification by transforming a complicated function into one that is easier to integrate.
• Integration by Parts is sometimes useful, especially when solving integrals involving products of functions, though not directly used in this problem.

The exercise employs substitution as the primary method to simplify the integral of exponential functions. For the final evaluation, direct integration using basic exponential rules completes the process. Familiarity with a range of methods enhances flexibility and problem-solving skills.

Exponential Functions

Exponential functions appear frequently in calculus, often characterized by having the base \( e \), a constant approximately equal to 2.718. In integration, exponential functions maintain their structure, which makes them somewhat easier to integrate compared to polynomials.
In the exercise, exponential functions are a key part of the integrand \( e^{x^2+y^2} \). Understanding how to handle these functions is crucial when you are integrating:

• The integral \( \int e^{u} \, du = e^{u} + C \) is a standard formula used extensively.
• Exponents inside an integral may require a substitution to handle them during integration, as demonstrated.

Exponential functions maintain their property during differentiation and integration, meaning the derivative and integral of \( e^x \) are \( e^x \), with minor adjustments involving coefficients and substitution. This property simplifies the integration process and was utilized effectively in solving the double integral problem.