91Ó°ÊÓ

A key concept when working with double integrals is the idea of an iterated integral. In an iterated integral, we evaluate a double integral by first performing integration with respect to one variable, and then subsequently integrating the result with respect to the second variable. This is particularly useful when dealing with regions that are described by specific bounds.
For instance, let's consider the region defined by

• \(0 \leq x \leq \sqrt{\pi/2}\)
• \(0 \leq y \leq 1\)

In this exercise, we rewrite the original double integral as an iterated integral:\[\int_{0}^1 \int_{0}^{\sqrt{\pi / 2}} x y \sin x^{2} \, dx \, dy\]This transformation allows us to tackle the integration process step-by-step, simplifying the approach by dealing with one variable at a time. Iterated integrals are essential in calculus to handle more complex integration tasks systematically.

Integration techniques are a fundamental component of calculus, enabling us to solve integrals that might otherwise be complex or impossible to tackle directly. In this context, after setting up the iterated integral, we first perform integration with respect to \(x\).
This requires finding the antiderivative of the expression \(x y \sin x^{2}\).
When approaching such integrals, typical techniques include

• Integration by parts
• Partial fractions
• Substitution methods

By applying these techniques effectively, we can transform difficult integrals into simpler forms.
Here, we combine substitution, a technique allowing variable replacement to simplify the integrand, with understanding trigonometric integrals, ultimately integrating first with \(x\) and then handling \(y\) with straightforward integration.

The substitution method is a strategic approach for simplifying integrals by changing variables.
In our exercise, we focus on the integral of the form \(x y \sin x^{2}\). This cannot be integrated directly with respect to \(x\) because of the function \(\sin x^{2}\).
Therefore, a substitution makes the process manageable. Setting \(u = x^2\), we differentiate to find

• \( \frac{du}{dx} = 2x\)
• Thus, \(dx = \frac{du}{2x}\)

This substitution transforms the integral to:\[y\int_{0}^{\pi / 2} \sin u \frac{du}{2}\]As a result, the expression becomes much simpler to integrate, allowing us to find the antiderivative, specifically with respect to \(u\). This method is vital in calculus, helping to simplify and solve expressions that involve complicated compositions of functions.