Question

Evaluate each of the iterated integrals. $$ \int_{0}^{\pi} \int_{0}^{1} x \sin y d x d y $$

Short answer

The value of the iterated integral is 1.

Step-by-step solution

Understand the Problem

We have an iterated integral to evaluate: \( \int_{0}^{\pi} \int_{0}^{1} x \sin y \, dx \, dy \). The first integration with respect to \(x\) is from 0 to 1, and the second integration with respect to \(y\) is from 0 to \(\pi\).

Inner Integration with Respect to x

Start by focusing on the inner integral \( \int_{0}^{1} x \sin y \, dx \). Treat \( \sin y \) as a constant with respect to \(x\). The integral becomes \( \sin y \int_{0}^{1} x \, dx \).

Evaluate the Inner Integral

Calculate \( \int_{0}^{1} x \, dx \). The antiderivative of \(x\) is \( \frac{x^2}{2} \). Evaluate this from 0 to 1: \( \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \).

Combine with Constant \( \sin y \)

Combine the result of the inner integral with \( \sin y \): \( \sin y \times \frac{1}{2} = \frac{1}{2} \sin y \).

Outer Integration with Respect to y

Perform the outer integral \( \int_{0}^{\pi} \frac{1}{2} \sin y \, dy \). Factor out the constant \( \frac{1}{2} \) to simplify: \( \frac{1}{2} \int_{0}^{\pi} \sin y \, dy \).

Evaluate the Outer Integral

Calculate \( \int_{0}^{\pi} \sin y \, dy \). The antiderivative of \( \sin y \) is \( -\cos y \). Evaluate it from 0 to \(\pi\): \( \left[-\cos y\right]_{0}^{\pi} = -\cos(\pi) + \cos(0) = 1 + 1 = 2 \).

Combine and Simplify

Combine the result with the constant: \( \frac{1}{2} \times 2 = 1 \).

Conclusion

The value of the original iterated integral is 1.

Key concepts

Integration Techniques

Integration techniques are methods used to find integrals, which are essential for solving many problems in calculus. There are a variety of techniques, each suited to different types of functions. In the context of iterated integrals, like the one given here, we often use techniques such as:

• Substitution: Useful for simplifying integrals by changing variables.
• Integration by parts: Helpful for integrals of products of functions.
• Recognizing derivatives: Identifying functions whose derivatives are easily integrable.

For iterated integrals, we decompose the problem into inner and outer integrals, each with respect to a different variable. First, we perform the inner integration, treating all other variables as constants. Then, the resulting expression is used in the outer integration. This layered approach is essential to tackling multi-variable integrals effectively.

Antiderivative

Antiderivatives are functions that represent the opposite of the derivative process. In other words, if you differentiate an antiderivative, you get the original function back. In the given solution, we frequently determine antiderivatives:For example, in the exercise:

• The antiderivative of \(x\) with respect to \(x\) is \(\frac{x^2}{2}\). This is because the derivative of \(\frac{x^2}{2}\) is \(x\).
• The antiderivative of \(\sin y\) with respect to \(y\) is \(-\cos y\). Differentiating \(-\cos y\) gives \(\sin y\).

Understanding antiderivatives is crucial for evaluating definite integrals, as you'll need them to compute the results at the integration limits.

Definite Integral

A definite integral is an integral with set upper and lower limits, providing a numerical value representing the area under a curve between these points. In the example, we computed definite integrals twice:

• The integral from 0 to 1 with respect to \(x\), represented as \([\frac{x^2}{2}]_{0}^{1}\), which results in \(\frac{1}{2}\).
• The integral from 0 to \(\pi\) with respect to \(y\), expressed as \([-\cos y]_{0}^{\pi}\), yielding 2.

Definite integrals yield specific values (like 1 in the final step). They are widely used in physics and engineering, where calculating exact amounts or changes over time is necessary. They are often resolved by first finding an antiderivative and then applying the limits, as shown in the exercise.

Integration Order

The order of integration in an iterated integral can significantly affect how challenging the problem is. It dictates the sequence in which integrations are performed, and each order will yield the same result, though some paths might be simpler than others.In our task, we first integrate with respect to \(x\) and then \(y\). Here's why it matters:

• Choosing to integrate \(x\) first makes \(\sin y\) act as a constant, simplifying the initial integration.
• After the inner calculation, the integral turns into an expression only involving \(y\), making the outer integration manageable.

In more complex problems, reversing the order of integration or choosing wisely based on function properties can simplify calculations dramatically. Practicing different integration orders helps in identifying the most efficient method for solving specific iterated integrals.