Problem 35 - Thomas Calculus

Question

In Exercises \(31-40,\) sketch the region of integration, reverse the order of integration, and evaluate the integral. $$ \int_{0}^{2 \sqrt{\ln 3}} \int_{y / 2}^{\sqrt{\ln 3}} e^{x^{2}} d x d y $$

Step-by-Step Solution

Answer

The integral is reversed and given by: \(\int_{0}^{\sqrt{\ln 3}} \int_{2x}^{2\sqrt{\ln 3}} e^{x^{2}} dy dx\) and needs numerical evaluation for precise value.

Step 1: Identify the Region of Integration

The given region is defined by the limits of the integral in terms of variables y and x: \ \(0 \leq y \leq 2\sqrt{\ln 3}\) and for each y, \(\frac{y}{2} \leq x \leq \sqrt{\ln 3}\). \ Plot these inequalities to sketch the region: \ \(x = y/2\) (a line from (0,0)), \(x = \sqrt{\ln 3}\) (a vertical line), and \(y = 2\sqrt{\ln 3}\) (horizontal line from x-axis to (\(\sqrt{\ln 3}, 2\sqrt{\ln 3}\)). \ This forms a triangular region in the xy-plane.

Step 2: Reverse the Order of Integration

To reverse the order of integration, we express y in terms of x. From the region, the x-boundaries are \(0\) to \(\sqrt{\ln 3}\). For each fixed x, y ranges from \(2x\) to \(2\sqrt{\ln 3}\) due to \(y = 2x\) and \(y = 2\sqrt{\ln 3}\). \ Therefore, the new integral is: \ \(\int_{0}^{\sqrt{\ln 3}} \int_{2x}^{2\sqrt{\ln 3}} e^{x^{2}}\, dy \, dx\).

Step 3: Evaluate the Inner Integral

Calculate the inner integral with respect to y: \ \(\int_{2x}^{2\sqrt{\ln 3}} e^{x^{2}} \, dy = e^{x^{2}} [y]_{2x}^{2\sqrt{\ln 3}}\). \ Substitute the limits: \ \(e^{x^{2}} (2\sqrt{\ln 3} - 2x)\).

Step 4: Evaluate the Outer Integral

Now evaluate the outer integral with respect to x: \ \(\int_{0}^{\sqrt{\ln 3}} e^{x^{2}} (2\sqrt{\ln 3} - 2x) \, dx\). \ This becomes two separate integrals: \ \(2\sqrt{\ln 3} \int_{0}^{\sqrt{\ln 3}} e^{x^{2}} \, dx - 2 \int_{0}^{\sqrt{\ln 3}} xe^{x^{2}} \, dx\).

Step 5: Solve Each Integral Separately

For the first integral: \ \( \int e^{x^{2}} \, dx\) does not have a simple antiderivative, so focus on special functions or numerical methods for evaluation.For the second integral: \ Use substitution \( u = x^2 \), \( du = 2x \, dx \). Thus, it simplifies to: \ \( - \int_{0}^{\ln 3} e^{u} \, du = - [e^{u}]_{0}^{\ln 3} = - (\ln 3 - 1) = 1 - \ln 3\).

Step 6: Combine Results

Substitute the evaluated integrals into the expression: \ \(2\sqrt{\ln 3} \int_{0}^{\sqrt{\ln 3}} e^{x^{2}} \, dx - 2(1 - \ln 3)\). \ The integral \(\int_{0}^{\sqrt{\ln 3}} e^{x^{2}} \, dx\) is left unevaluated analytically. Approximate using numerical methods or refer to tabulated values to evaluate this integral accurately. Combine this result with the other term for a final numerical answer.

Key Concepts

Double Integrals

In multivariable calculus, double integrals are a way of calculating the volume under a surface over a specific region in the plane. The integral \[\int_{a}^{b} \int_{c}^{d} f(x, y) \, dx \, dy\]represents summing small infinite pieces over a defined area. What makes double integrals powerful is their ability to handle functions of two variables, which are often encountered in physics and engineering. They are typically evaluated by computing the integral with respect to one variable while considering the other variable as a constant, and then repeating the process for the second variable. This step-by-step method allows you to find the sum of all the infinitely small pieces that make up the volume under the surface.

• Imagine slicing a loaf of bread into infinitely thin pieces - double integrals do that mathematically!
• They are part of how we find multidimensional 'areas', like 3D volumes.

Recognizing the structure and limits of integration is key to applying double integrals effectively.

Region of Integration

The region of integration in a double integral is the area over which the integration is performed. This region is defined by the boundaries or limits of the integral and is crucial because it informs us where to evaluate the function. In our exercise, the region is determined by inequalities involving x and y: \[\begin{align*}0 & \leq y \leq 2\sqrt{\ln 3}, \\frac{y}{2} & \leq x \leq \sqrt{\ln 3}.\end{align*}\] It's helpful to sketch this region to visualize the triangle in the xy-plane. Key elements to note:

• Boundary lines like \(x = y/2\) act as borders for the region.
• Limits such as \(x = \sqrt{\ln 3}\) and \(y = 2\sqrt{\ln 3}\) help define the maximal and minimal values within which the function is evaluated.

Understanding the region of integration helps ensure the results of a double integral correctly represent the desired area.

Order of Integration

Reversing the order of integration means swapping the variables with respect to which we integrate first and second. This is often necessary to simplify the computation. In our example, reversing changes the order from \(\int (\text{with } y) \int (\text{with } x)\) to \(\int (\text{with } x) \int (\text{with } y)\). We change the limits accordingly, describing the region using different boundaries: This changes from \(\int_{0}^{2\sqrt{\ln 3}} \int_{y/2}^{\sqrt{\ln 3}} \) to \(\int_{0}^{\sqrt{\ln 3}} \int_{2x}^{2\sqrt{\ln 3}}.\)Reversing the order can:

• Make an otherwise difficult integral tractable.
• Align the integration with known functions that are easier to evaluate.

Choosing the right order is often a critical step in simplifying the solution to mathematical problems involving multiple integrals.

Numerical Integration

There are instances, especially with complex functions, where analytical solutions for integrals are elusive. Numerical integration is a technique used to approximate the value of an integral, particularly useful when dealing with integrals without simple antiderivatives, like \(\int e^{x^{2}} \, dx.\) Methods such as Simpson's Rule, the Trapezoidal Rule, or numerical libraries in software can approximate these integrals.

• Computer software can quickly compute these approximations, providing practical solutions when exact answers aren't necessary.
• Tabulated values for special functions, often used in scientific fields, help find approximate integrals.

In all, numerical integration is indispensable in circumstances where functions are complex or transcendental, efficiently offering a route to approximate solutions.