Problem 49 - Thomas Calculus

Question

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \begin{equation} \int_{0}^{1} \int_{y}^{1} x^{2} e^{x y} d x d y \end{equation}

Step-by-Step Solution

Answer

The final value of the integral is \( \frac{1}{2}e - 1 \).

Step 1: Understand the Limits of Integration

The given integral is \( \int_{0}^{1} \int_{y}^{1} x^{2} e^{xy} \, dx \, dy \). The inner integral \( \int_{y}^{1} \) corresponds to the variable \( x \), and it ranges from \( y \) to \( 1 \). The outer integral \( \int_{0}^{1} \) corresponds to \( y \), which ranges from \( 0 \) to \( 1 \). Therefore, the region of integration is defined as where \( x \) ranges between \( y \) and \( 1 \), and \( y \) ranges from \( 0 \) to \( 1 \).

Step 2: Sketch the Region of Integration

To visualize the region, draw a coordinate plane with axes \( x \) and \( y \). The region \( x = y \) to \( x = 1 \) starts on the line \( y=x \) and goes to the vertical line \( x=1 \). The region is bounded below by the line \( y=0 \). Thus, the area is a right triangle with vertices at \( (0, 0) \), \( (1, 1) \), and \( (1, 0) \).

Step 3: Reverse the Order of Integration

To reverse the order of integration, switch the order of \( dx \) and \( dy \). In the described triangle, \( y \) varies horizontally from \( 0 \) to the line \( x \), and \( x \) varies from \( 0 \) to \( 1 \). Therefore, the new limits are \( \int_{0}^{1} \int_{0}^{x} x^{2} e^{xy} \, dy \, dx \).

Step 4: Compute the Integral with Reversed Limits

Now, compute \( \int_{0}^{1} \int_{0}^{x} x^{2} e^{xy} \, dy \, dx \). First integrate with respect to \( y \):\[ \int_{0}^{x} x^{2} e^{xy} \, dy = x^{2} \left[ \frac{e^{xy}}{x} \right]_{0}^{x} = x^{2} \left( \frac{e^{x^2} - e^{0}}{x} \right) = x(e^{x^2} - 1). \]Then integrate the result with respect to \( x \):\[ \int_{0}^{1} x(e^{x^2} - 1) \, dx. \]

Step 5: Evaluate the Final Integral

Split the integral: \( \int_{0}^{1} xe^{x^2} \, dx - \int_{0}^{1} x \, dx \). For the first part, use substitution: let \( u = x^2 \), thus \( du = 2x \, dx \), so \( x \, dx = \frac{1}{2} du \):\[ \int xe^{x^2} \, dx = \frac{1}{2} \int e^{u} \, du = \frac{1}{2} e^{u} + C. \]Evaluate from 0 to 1, resulting in \( \frac{1}{2}(e - 1) \). The second integral is simpler: \( \int x \, dx = \frac{x^2}{2} \), from 0 to 1, resulting in \( \frac{1}{2} \). The evaluated integral is \( \frac{1}{2}(e - 1) - \frac{1}{2} = \frac{1}{2}e - 1 \).

Key Concepts

Reversing Order of Integration

Reversing the order of integration in a double integral can simplify complex calculations. This is particularly helpful when the region of integration is easier to handle with a different order. Consider our integral \( \int_{0}^{1} \int_{y}^{1} x^{2} e^{xy} \, dx \, dy \). Here, the inner integral is over \(x\) from \(y\) to \(1\), while \(y\) ranges from \(0\) to \(1\). By reversing this order, we switch to \( \int_{0}^{1} \int_{0}^{x} x^{2} e^{xy} \, dy \, dx \). This new setup starts \(y\) from \(0\) to \(x\), and \(x\) from \(0\) to \(1\). This can dramatically affect the ease of solving the integral.

Region of Integration

The region of integration is a crucial aspect of understanding double integrals. It defines the area over which the integration is performed. In our task, the region is a right triangle on the \(xy\)-plane bounded by the lines \(y=x\), \(x=1\), and \(y=0\). Specifically, \(x\) progresses from \(y\) to \(1\), and \(y\) spans from \(0\) to \(x\). Understanding this triangular region helps in determining the limits of integration when reversing the order, or even when performing the initial integral computation. It visually clarifies where one variable limits the other.

Substitution Method

The substitution method is a valuable integral solving technique that simplifies parts of an expression by substituting variables. When solving \( \int xe^{x^2} \, dx \), let \( u = x^2 \). Thus, \( du = 2x \, dx \), making \( x \, dx = \frac{1}{2} du \). This substitution transforms the integral into \( \frac{1}{2} \int e^u \, du \), which is straightforward to solve, turning otherwise complex expressions into manageable forms. The key to using substitution is identifying the part of the integral that can be conveniently replaced and differentiated.

Integration by Parts

Integration by parts is a strategy based on the product rule for differentiation, particularly useful for integrals involving products of functions. To evaluate \( \int x e^{x^2} \, dx \) using integration by parts might be less straightforward due to the complexity compared to direct substitution. However, the formula \( \int u \, dv = uv - \int v \, du \) is the backbone. In cases where substitution isn't suitable, split the integral into products of easily integrable functions. This strategy proves invaluable in a wide range of integrals, breaking them into simpler, solvable parts.