Problem 21 - Thomas Calculus

Question

Sketch the region of integration and evaluate the integral. \begin{equation} \int_{1}^{\ln 8} \int_{0}^{\ln y} e^{x+y} d x d y \end{equation}

Step-by-Step Solution

Answer

The value of the integral is \( 8 \ln 8 - 15 \).

Step 1: Understand the Limits of Integration

The given integral is a double integral \( \int_{1}^{\ln 8} \int_{0}^{\ln y} e^{x+y} \, dx \, dy \). The outer integral \( \int_{1}^{\ln 8} \, dy \) indicates that \( y \) ranges from 1 to \( \ln 8 \), and for each fixed \( y \), the inner integral \( \int_{0}^{\ln y} \, dx \) indicates that \( x \) ranges from 0 to \( \ln y \).

Step 2: Sketch the Region of Integration

To sketch the region, start by drawing the \( xy \)-plane. The region is bounded vertically by \( y = 1 \) and \( y = \ln 8 \). For each \( y \), \( x \) ranges from \( x = 0 \) to \( x = \ln y \). This creates a region bounded by the horizontal lines \( y = 1 \), \( y = \ln 8 \), the vertical line \( x = 0 \), and the curve \( x = \ln y \) which is \( y = e^x \).

Step 3: Set Up the Integral with Exponential Function

The integrand is \( e^{x+y} \). We start evaluating the inner integral with respect to \( x \), keeping \( y \) constant: \( \int_{0}^{\ln y} e^{x+y} \, dx \).

Step 4: Perform the Inner Integration

Using the substitution \( u = x + y \), such that \( du = dx \), for the inner integral \( \int_{0}^{\ln y} e^{x+y} \, dx \), calculate the antiderivative: \( \int e^u \, du = e^u \). Therefore, evaluate from 0 to \( \ln y \): \( e^{x+y} \bigg|_0^{\ln y} = e^{\ln y + y} - e^{y} = y e^y - e^y = (y - 1) e^y \).

Step 5: Perform the Outer Integration

Now, integrate the result \((y - 1) e^y\) with respect to \( y \) over the range from 1 to \( \ln 8 \): \( \int_1^{\ln 8} (y - 1) e^y \, dy \). Use integration by parts: let \( u = y - 1 \), \( dv = e^y \, dy \); then \( du = dy \), \( v = e^y \). The integration by parts formula \( \int u \, dv = uv - \int v \, du \) gives \( (y-1)e^y \bigg|_1^{\ln 8} - \int_1^{\ln 8} e^y \, dy \).

Step 6: Compute Integration by Parts

Evaluate the expression: \((y-1)e^y -(e^y)\) from \( 1 \) to \( \ln 8 \) gives: \( ((\ln 8 - 1) e^{\ln 8} - e^{\ln 8}) - ((1-1)e^1 - e^1) = 8 (\ln 8 - 1) - 8 - (-1)\).

Step 7: Simplify and Calculate the Result

Simplify the expression: \( 8 \ln 8 - 8 - 8 + 1 = 8 \ln 8 - 15 \). Thus, the value of the integral is \( 8 \ln 8 - 15 \).

Key Concepts

Region of Integration

A double integral involves integrating over a specific region in the coordinate plane, and understanding this region is crucial to solving related problems. Let's break down how to determine this region. In our example, the integral given is \( \int_{1}^{\ln 8} \int_{0}^{\ln y} e^{x+y} \, dx \, dy \). The limits provided by the integrals delineate a region in the xy-plane.

• The outer integral, \( \int_{1}^{\ln 8} \, dy \), indicates that y ranges from 1 to \( \ln 8 \).
• For each fixed value of y in this interval, x varies from 0 to \( \ln y \), as per the inner integral, \( \int_{0}^{\ln y} \, dx \).

To visualize this region, you can sketch the xy-plane. Draw horizontal lines \( y = 1 \) and \( y = \ln 8 \) to show the vertical limits. For each y-value, consider that x ranges between \( x = 0 \) and \( x = \ln y \), creating a boundary where any point lies below the curve \( x = \ln y \), which corresponds to \( y = e^x \). Together, these limits form a curved trapezoidal region.

Exponential Function

Exponential functions play a central role in many mathematical contexts, particularly in calculus. Here, the function \( e^{x+y} \) is the integrand in our integral. Let's dive deeper into its properties.

• The exponential function, denoted as \( e^u \), is notable for its rapid rate of growth compared to polynomial functions.
• It possesses an important property: the derivative of \( e^u \) is itself \( e^u \), which makes it straightforward yet powerful in calculus.

In our example, \( e^{x+y} \) suggests the need to integrate a function that grows exponentially with the sum of x and y. During the integration process, recognizing that \( e^{x+y} = e^x \cdot e^y \) allows us to approach the integral more systematically. This breakdown highlights how exponential functions facilitate solving complex integrals due to their unique characteristics.

Integration by Parts

Integration by Parts is a technique derived from the product rule of differentiation and is invaluable in solving integrals. It is particularly useful when the integrand is a product of two functions.

• The method is based on the formula: \( \int u \, dv = uv - \int v \, du \), where \( u \) and \( dv \) are well-chosen parts of the integrand.
• Choosing \( u \) such that its derivative \( du \) simplifies the problem, while \( dv \) should be easily integrable, is strategic.

In our exercise, we apply integration by parts to \( (y - 1) e^y \). Here:

• Let \( u = y - 1 \) and \( dv = e^y \, dy \); consequently, \( du = dy \) and \( v = e^y \).
• Applying the formula, we find \( (y-1)e^y - \int e^y \, dy \).

This simplifies to evaluating boundary terms and the exponential function, leading to a concise outcome. Mastering Integration by Parts allows leaps in problem-solving capabilities for integrals involving products of polynomial and exponential terms.