Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals. $$ \int_{0}^{\pi / 2} \int_{0}^{\sin y} e^{x} \cos y d x d y $$

Step-by-Step Solution

Answer

The iterated integral evaluates to \( e - 2 \).

Step 1: Understand the problem

We are given a double integral \( \int_{0}^{\pi / 2} \int_{0}^{\sin y} e^{x} \cos y \ dx \ dy \). We need to evaluate this integral by first integrating with respect to \( x \), and then integrating the result with respect to \( y \).

Step 2: Integrate with respect to x

For the integral \( \int_{0}^{\sin y} e^{x} \cos y \ dx \), we treat \( \cos y \) as a constant with respect to \( x \). The integral of \( e^x \) with respect to \( x \) is \( e^x \). Hence, the integral becomes:\[\int_{0}^{\sin y} e^{x} \cos y \ dx = \left. e^x \cos y \right|_0^{\sin y} = e^{\sin y} \cos y - e^0 \cos y = (e^{\sin y} - 1) \cos y,\]since \( e^0 = 1 \).

Step 3: Integrate with respect to y

Now, substitute the result from Step 2 into the outer integral:\[\int_{0}^{\pi / 2} (e^{\sin y} - 1) \cos y \ dy.\]This can be split into two separate integrals:\[\int_{0}^{\pi / 2} e^{\sin y} \cos y \ dy - \int_{0}^{\pi / 2} \cos y \ dy.\]For the first integral, use the substitution \( u = \sin y \), \( du = \cos y \ dy \), changing the limits of integration accordingly. When \( y = 0 \), \( u = 0 \), and when \( y = \pi/2 \), \( u = 1 \). The integral becomes:\[\int_{0}^{1} e^u \ du = \left. e^u \right|_0^1 = e^1 - e^0 = e - 1.\]For the second integral, \( \int_{0}^{\pi / 2} \cos y \ dy = \left. \sin y \right|_0^{\pi / 2} = 1 - 0 = 1. \)

Step 4: Combine the results

Subtract the results of the second integral from the result of the first integral:\[(e - 1) - 1 = e - 2. \]

Key Concepts

Double Integral

A double integral is an advanced calculus operation used to calculate the volume under a surface over a two-dimensional region. In simple terms, it's like stacking a bunch of tiny shapes on top of each other to measure the total space they occupy.
The integral \[\int \int_R f(x, y) \, dx \, dy\]
is evaluated over a specified region \( R \) in the \( xy \)-plane. Here, we are splitting the calculation into two steps:

• The inner integral is solved first (with respect to one variable, while treating other variables as constants).
• The result is then integrated with respect to the second variable via the outer integral.

Double integrals are especially useful in physics and engineering, allowing calculation of areas, volumes, and averages over a region. This specific exercise involves calculating an iterated integral to find the solution.

Iterated Integrals

Iterated integrals are a way to evaluate double integrals by solving them step-by-step. Instead of handling both variables at once, we tackle one at a time.
In our exercise, we've been given:\[\int_{0}^{\pi / 2} \int_{0}^{\sin y} e^{x} \cos y \ dx \ dy \]

• The inner integral:
  Here, the initial step is to integrate with respect to \( x \), treating \( \cos y \) as a constant. This simplifies our calculations since the derivative of \( e^x \) with respect to \( x \) stays \( e^x \).
• The outer integral:
  With the result from the inner integration step, we integrate with respect to \( y \), often simplifying further by substitutions, as required.

Iterated integrals make double integrals easier to handle, enabling us to solve complex problems by splitting the task into manageable parts.

Substitution Method

The substitution method is a handy technique to make integrals easier by transforming variables. Think of it as changing coordinates to simplify the problem.
In the given exercise, once we had the result of the inner integral, we needed to handle:
\[\int_{0}^{\pi / 2} (e^{\sin y} - 1) \cos y \ dy\]
For the first part of this integral:

• We used a substitution where \( u = \sin y \), leading to \( du = \cos y \, dy \).
• Changing integral limits accordingly made integration simpler:
  When \( y = 0 \), \( u = 0 \). When \( y = \pi/2 \), \( u = 1 \).
  This transformed the first integral into \( \int_{0}^{1} e^u \, du \).

Using the substitution method, we simplify calculations, making complex integrals more straightforward. It’s a helpful tool in calculus to transform and evaluate integrals effectively.

Integral Calculus

Integral calculus is the mathematical study of accumulation and area. It revolves around the concept of integration, used to compute quantities such as areas under curves, volumes, and more.
In this problem, integral calculus is fundamental. We apply it using:

• Double integrals: To compute the area beneath a surface (or volume beneath a surface) by evaluating a function over a described region.
• Iterated integrals: Breaking down the double integral into two smaller tasks, to simplify and handle each variable separately.
• Substitution method: Transforming the variables to make difficult integrals easier to evaluate.

The overall aim in integral calculus through exercises like our given problem is to understand how to effectively use integrals to solve practical problems related to volume and accumulation across various fields of study.