Problem 23 - Calculus Early Transcendentals

Question

Evaluate each double integral over the region \(R\) by converting it to an iterated integral. $$\iint_{R} e^{x+2 y} d A ; R=\\{(x, y): 0 \leq x \leq \ln 2,1 \leq y \leq \ln 3\\}$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral of \( \displaystyle \iint_R e^{x + 2y} \, dA \), where \(R = \{ (x, y) : 0 \leq x \leq \ln2, 1 \leq y \leq \ln3\} \). Answer: \( \frac{1}{2}(9 - e^2) \).

Step 1: Observe the region of integration

We are given the region of integration \(R\) as: \({0 \leq x \leq \ln 2,1 \leq y \leq \ln 3}\). This is a rectangular region with \(x\) varying from \(0\) to \(\ln 2\), and \(y\) varying from \(1\) to \(\ln 3\).

Step 2: Write the double integral as an iterated integral

We can write the given double integral as an iterated integral in which we first integrate with respect to \(x\) and then with respect to \(y\). $$\iint_{R} e^{x+2 y} d A=\int_{1}^{\ln3}\int_{0}^{\ln2} e^{x+2 y} d x d y$$

Step 3: Integrate with respect to \(x\)

Now, we can integrate the inner integral with respect to \(x\): $$\int_{0}^{\ln2} e^{x+2 y} d x = \left[e^{x+2 y}\right]_{0}^{\ln2} = e^{\ln2+2 y} - e^{0+2 y} = (e^{\ln2} - 1)e^{2y}=(2 - 1)e^{2y} = e^{2y}$$

Step 4: Integrate with respect to \(y\)

Now, we can continue by integrating the outer integral with respect to \(y\): $$\int_{1}^{\ln3} e^{2 y} d y = \frac{1}{2}\left[e^{2 y}\right]_{1}^{\ln3} = \frac{1}{2}(e^{2\ln3} - e^{2(1)}) = \frac{1}{2}(e^{\ln 9} - e^{2}) = \frac{1}{2}(9 - e^{2})$$

Step 5: Final answer

Therefore, the value of the given double integral is \(\frac{1}{2}(9 - e^{2})\).

Key Concepts

Understanding Integration

Integration is a fundamental concept in calculus which involves finding the area under a curve or the accumulation of quantities. In single-variable calculus, when you integrate a function, you're essentially summing up an infinite number of infinitesimally small quantities to find cumulative values, such as areas or total distances.
For instance, if you have a curve defined by a function \(f(x)\), the definite integral \(\int_{a}^{b} f(x) \, dx\) calculates the total area under the curve from \(x = a\) to \(x = b\). When dealing with functions of two variables, like \(f(x, y)\), the concept of integration extends to double integrals. These are used to find volumes under surfaces over a given region in the xy-plane. In such cases, the double integral \(\iint_{R} f(x, y) \, dA\) computes the volume under a surface over the region \(R\). The region \(R\) can often be a rectangle, triangle, or other defined plane areas. To perform these calculations, the double integral is converted into an iterated integral, simplifying the integration into separate steps for each variable.

What is an Iterated Integral?

An iterated integral helps us evaluate double integrals by breaking them into a sequence of simple integrals. This process involves integrating a function with respect to one variable first, and then using the result to integrate with respect to the other variable. In our exercise, we first integrate \(e^{x+2y}\) with respect to \(x\), while keeping \(y\) as a constant, to compute the inner integral.
We take the solution from this integration and then integrate it with respect to \(y\), which is known as the outer integral.The order of integration (whether \(x\) or \(y\) first) can sometimes be swapped depending on the region of integration, but it must follow the limits of integration according to the problem's specifications. In rectangular regions, this involves integrating \(x\) over its limits and then \(y\), or vice versa. Iterated integrals simplify the double integral process into more manageable single-integral calculations.
When dealing with complex functions, iterated integrals provide a step-by-step pathway to solve the integral neatly.

Rectangular Region in Double Integrals

The concept of a rectangular region is straightforward and commonly used in double integrals. A rectangular region is defined by constant upper and lower limits for both variables.
In this exercise, the region \(R\) is given by the boundaries: \(0 \leq x \leq \ln 2\), and \(1 \leq y \leq \ln 3\). The simplicity of rectangular regions allows the limits of integration to be clearly defined, making it easier to arrange the iterated integrals. For example, we can set up our iterated integral as \(\int_{1}^{\ln 3} \int_{0}^{\ln 2} e^{x+2y} \, dx \, dy\), where the inner integral \(\int_{0}^{\ln 2} e^{x+2y} \, dx\) has set limits for \(x\) between 0 and \(\ln 2\), and the outer integral \(\int_{1}^{\ln 3} \, dy\) has \(y\) varying between 1 and \(\ln 3\). Such regions make it straightforward to apply integration with respect to each variable systematically.
In cases where the limits vary with respect to the variables, solving the double integral might require altering the order of integration or changing integration strategies. Rectangular regions usually simplify this process, as the order of integration remains consistent and does not depend on the other variable, except for maintaining the function's correct evaluation.