Problem 33 - Calculus an Applied Approach

Question

In Exercises 33 and \(34,\) evaluate the double integral. Note that it is necessary to change the order of integration. $$ \int_{0}^{3} \int_{y}^{3} e^{x^{2}} d x d y $$

Step-by-Step Solution

Answer

\(\frac{e^{9} - 1}{2}\)

Step 1: Find the common range of \(x\) and \(y\)

First, look at the bounds of the initial integral: \(0 \leq y \leq 3\) and \(y \leq x \leq 3\). Then sketch these inequalities on a coordinate plane. From the diagram, it can be inferred that the common range is \(0 \leq y \leq x \leq 3\).

Step 2: Change the order of integration

The order of the old integral is \(dydx\), but we need to change it to \(dxdy\). The new limits should be found out by considering the region of \(x\) and \(y\). It is clear that \(x\) varies from 0 to 3 and \(y\) varies from 0 to \(x\). So, the new double integral is \( \int_{0}^{3} \int_{0}^{x} e^{x^{2}} dy dx \) .

Step 3: Evaluate the inner integral

The inner integral is in terms of \(y\), but there are no \(y\)'s in the integrand, so the \(y\) just acts as a scalar multiplier. Therefore, we have \( \int_{0}^{3} [ ye^{x^2} ]_0^x dx = \int_{0}^{3} x e^{x^{2}} dx \).

Step 4: Evaluate the outer integral

Now, we let \(u = x^2\). Then \(du = 2x dx\). Our integral now becomes a simple integral of \(e^u\). Now the integral can be rewritten as: \( \frac{1}{2} \int e^{u} du = \frac{1}{2} e^{u} = \frac{1}{2} e^{x^2}\). The evaluated outer integral from 0 to 3 is then \( \frac{1}{2} e^{x^2} ]_0^3 = \frac{e^{9}}{2} - \frac{1}{2} = \frac{e^{9} - 1}{2} \).

Key Concepts

Change of Integration Order

In solving double integrals, it may sometimes be necessary to change the order of integration. This process is crucial if the integration limits do not allow for straightforward computation or when simplifying the integration process. We start with the initial problem setup:

• The original integral: \( \int_{0}^{3} \int_{y}^{3} e^{x^{2}} \, dx \, dy \)
• The integration is first performed with respect to \( x \), then \( y \).

To change the order from \( dxdy \) to \( dydx \), we need to understand how the variables limit each other in the given domain. The limits for \( y \) can be read as \( 0 \leq y \leq x \), and for \( x \) as \( y \leq x \leq 3 \).By sketching the region defined by these inequalities in the coordinate plane, we identify a triangular region where changing the order of integration will make the limits straightforward. The transformed integral becomes \( \int_{0}^{3} \int_{0}^{x} e^{x^2} \, dy \, dx \). This sets the stage for easier integration.

Integration Limits

Integration limits define the domain over which we integrate and directly affect the complexity of the evaluation. Initially, the limits are colinear over two intervals: \( 0 \leq y \leq 3 \) and \( y \leq x \leq 3 \). When changing the order of integration, new limits are constructed from these inequalities:

• For \( x \), it ranges from \( 0 \) to \( 3 \).
• Within these bounds, \( y \) varies from \( 0 \) to \( x \).

These new limits result in the integral: \( \int_{0}^{3} \int_{0}^{x} e^{x^2} \, dy \, dx \).The choice of integration limits determines the piece of the plane over which we'll perform the integration. Ensuring these limits are correct helps avoid errors and accurately defines the region of integration.

Evaluation of Integrals

The evaluation of integrals consists of performing the actual integration process as determined by the order and limits. Starting with the inner integral:

• Perform the inner integration \( \int_{0}^{x} e^{x^2} \, dy \).
• Since \( e^{x^2} \) is a constant with respect to \( y \), it multiplies by the width of the interval: \( x - 0 = x \).

This reduces to \( x \cdot e^{x^2} \).Next is the evaluation of the outer integral \( \int_{0}^{3} x \cdot e^{x^2} \, dx \). This requires a method called substitution.

U-Substitution

U-substitution is a technique used to simplify the integration process by making a substitution that changes the variable of integration. For our example:

• We set \( u = x^2 \), hence \( du = 2x \, dx \).
• This implies \( x \, dx = \frac{1}{2} du \).

Substituting these values in the integral, it becomes \( \frac{1}{2} \int e^u \, du \).The integration of \( e^u \) with respect to \( u \) is straightforward, resulting in \( \frac{1}{2} e^u \).Returning to the \( x \) variable, substitute back \( u = x^2 \) to get \( \frac{1}{2} e^{x^2} \). The final integration over the limits 0 to 3 leads to the expression \( \frac{e^{9} - 1}{2} \), yielding the evaluated result of the original problem.