Problem 51 - Thomas Calculus

Question

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \begin{equation} \int_{0}^{2 \sqrt{\ln 3}} \int_{y / 2}^{\sqrt{\ln 3}} e^{x^{2}} d x d y \end{equation}

Step-by-Step Solution

Answer

The integral simplifies to: \(2\sqrt{\ln 3} \int_{0}^{\sqrt{\ln 3}} e^{x^{2}} \ dx - 2\).

Step 1: Understand the Region of Integration

The given integral is \[ \int_{0}^{2 \sqrt{\ln 3}} \int_{y / 2}^{\sqrt{\ln 3}} e^{x^{2}} \ dx \ dy \]which means that for each value of \(y\) from 0 to \(2\sqrt{\ln 3}\), \(x\) will vary from \(y/2\) to \(\sqrt{\ln 3}\). This defines the region of integration in the xy-plane, bounded by the lines \(x = y/2\) and \(x = \sqrt{\ln 3}\), and between \(y = 0\) and \(y = 2\sqrt{\ln 3}\).

Step 2: Sketch the Region of Integration

To sketch the region, draw the line \(x = y/2\) and the vertical line \(x = \sqrt{\ln 3}\) on the xy-plane. The line \(y=2x\) intersects \(y=2\sqrt{\ln 3}\) at the point where \(x = \sqrt{\ln 3}\). The region is bounded below by \(y=0\), on the left by \(x = y/2\), and on the right by \(x = \sqrt{\ln 3}\), extending up to \(y=2\sqrt{\ln 3}\).

Step 3: Reverse the Order of Integration

To reverse the order, we need to express \(y\) in terms of \(x\). The bounds for \(x\) were originally from \(y/2\) to \(\sqrt{\ln 3}\). Now, for a fixed \(x\), we will have \(x = y/2\) or \(y = 2x\) as the lower bound, and the maximum \(y = 2\sqrt{\ln 3}\) as the upper bound. So, \(x\) will vary from 0 to \(\sqrt{\ln 3}\). The new limits for \(y\) are from \(2x\) to \(2\sqrt{\ln 3}\). Thus, the new integral becomes:\[ \int_{0}^{\sqrt{\ln 3}} \int_{2x}^{2\sqrt{\ln 3}} e^{x^{2}} \ dy \ dx \]

Step 4: Evaluate the New Integral

The new form of the integral is:\[ \int_{0}^{\sqrt{\ln 3}} \int_{2x}^{2\sqrt{\ln 3}} e^{x^{2}} \ dy \ dx \]First, integrate with respect to \(y\):\[ \int_{2x}^{2\sqrt{\ln 3}} e^{x^{2}} \ dy = e^{x^{2}}[y]_{2x}^{2\sqrt{\ln 3}} = e^{x^{2}}(2\sqrt{\ln 3} - 2x) \]The problem then simplifies to:\[ \int_{0}^{\sqrt{\ln 3}} e^{x^{2}}(2\sqrt{\ln 3} - 2x) \ dx \]Next, separate the integral:\[ 2\sqrt{\ln 3} \int_{0}^{\sqrt{\ln 3}} e^{x^{2}} \ dx - 2 \int_{0}^{\sqrt{\ln 3}} xe^{x^{2}} \ dx \]

Step 5: Solve Each Part of the Integral

For \( \int e^{x^{2}} \ dx \), it has no elementary antiderivative. Thus it is kept as is. But\[ \int_{0}^{\sqrt{\ln 3}} xe^{x^{2}} \ dx \]is solvable by substitution: setting \( u = x^{2} \), then \( du = 2x \ dx \), yields\[ \frac{1}{2} \int_{0}^{\ln 3} e^{u} \ du = \frac{1}{2} [e^{u}]_{0}^{\ln 3} = \frac{1}{2} (e^{\ln 3} - e^{0}) = \frac{1}{2} (3 - 1) = 1 \]The integral becomes simplified as:\[ 2\sqrt{\ln 3} \int_{0}^{\sqrt{\ln 3}} e^{x^{2}} \ dx - 2 \times 1 = 2\sqrt{\ln 3} \int_{0}^{\sqrt{\ln 3}} e^{x^{2}} \ dx - 2 \]

Step 6: Conclusion

After separate evaluation:- Without an elementary form for \(\int e^{x^{2}} \ dx\), it's typically evaluated using numerical integration or table values for specific limits.- The subtraction of the constant \(2\) is straightforward, resulting in our simplified result.

Key Concepts

Region of Integration

In the context of double integrals, the region of integration denotes the area over which the function is being integrated. Consider the integral \[\int_{0}^{2 \sqrt{\ln 3}} \int_{y / 2}^{\sqrt{\ln 3}} e^{x^{2}} \ dx \ dy\]This defines a specific region on the xy-plane. The variable \(y\) ranges from 0 to \(2\sqrt{\ln 3}\). For each fixed \(y\), \(x\) varies from \(y/2\) to \(\sqrt{\ln 3}\). Thus, the integration occurs over a region determined by these bounds.To visualize this, imagine sketching the region. You draw two lines, \(x = y/2\) and \(x = \sqrt{\ln 3}\). The first line is sloped, while the second is vertical. The region is also bounded by \(y=0\) and \(y=2\sqrt{\ln 3}\). This creates a confined area within the xy-plane, marking the region where integration is calculated.

Order of Integration

The order of integration in a double integral refers to the sequence in which integration is performed over different variables.For the integral \(\int_{0}^{2 \sqrt{\ln 3}} \int_{y / 2}^{\sqrt{\ln 3}} e^{x^{2}} \ dx \ dy\), the integration is first done with respect to \(x\) (inner integral) within limits \(y/2\) to \(\sqrt{\ln 3}\), followed by \(y\) (outer integral) from 0 to \(2\sqrt{\ln 3}\).Reversing the order involves expressing the bounds differently. For a fixed \(x\), we've redefined the integral as:\[\int_{0}^{\sqrt{\ln 3}} \int_{2x}^{2\sqrt{\ln 3}} e^{x^{2}} \ dy \ dx\]Here, \(x\) varies from 0 to \(\sqrt{\ln 3}\) with \(y\) varying between \(2x\) and \(2\sqrt{\ln 3}\). Changing the order can be necessary to ease integration, depending on the function and its bounds.

Exponential Functions

Exponential functions are functions where the variable appears as an exponent, such as \(e^{x^2}\) in our integration problem.Characteristics:

• Exhibit rapid growth or decay.
• Have the form \(e^{f(x)}\), where \(e\) is the base of natural logarithms, approximately equal to 2.718.
• Can be challenging when integrated, especially if it's a non-linear exponent like \(x^2\).

In this context, \(e^{x^2}\) lacks a simple antiderivative in elementary terms, meaning it cannot be expressed neatly using basic functions like polynomials or trigonometric functions. Numerical methods or series expansions are often used if more detailed evaluation is needed.

Integration by Substitution

Integration by substitution is a technique used to simplify an integral by changing variables. It's similar to reverse chain rule and is useful when faced with complex integrals.In the given problem, while integrating \(xe^{x^2}\), substitution eases the process:

• Set \(u = x^2\), leading to \(du = 2x\ dx\).
• This transforms the integral \(\int xe^{x^2} dx\) to \(\frac{1}{2} \int e^{u} du\).
• The integration in terms of \(u\) is straightforward, resulting in \(\frac{1}{2} (e^{u} )\).

The limits of integration change accordingly. Originally from \(x = 0\) to \(\sqrt{\ln 3}\), they become \(u = 0\) to \(u = \ln 3\), based on the substitution. This approach simplifies the integral from a difficult function into a more manageable form, demonstrating how variable substitution can be a powerful tool in calculus.