91Ó°ÊÓ

When working with iterated integrals, the first step involves evaluating the **inner integral**. In the given exercise, the inner integral is \( \int_{0}^{\sqrt{4-y^{2}}} x \, dx \). This integral requires us to find the antiderivative of the function \( x \) with respect to \( x \). The result is \( \frac{x^2}{2} \), a basic formula from calculus useful in many applications.
What makes this process interesting is that instead of using fixed limits, the upper limit itself, \( \sqrt{4-y^2} \), is an expression involving \( y \). This demands substitution once the antiderivative is found. When solving, plug in the limits:\[\left[ \frac{x^2}{2} \right]_{0}^{\sqrt{4-y^{2}}} = \frac{(\sqrt{4-y^{2}})^2}{2} - \frac{0^2}{2}\]This simplifies to \( \frac{4-y^2}{2} \).
Understanding and simplifying the inner integral is crucial as it affects all subsequent calculations. The outcome sets the stage for the outer integral, defining a new function in terms of \( y \) to be integrated next.

Once the inner integral is evaluated, we shift our focus to the **outer integral**. Here, it is \( \int_{0}^{2} \frac{4-y^2}{2} \, dy \). To simplify our work, we first express the integrand as \( 2 - \frac{y^2}{2} \).
The goal here is to integrate this new function with respect to \( y \) over the specified limits, \( y = 0 \) to \( y = 2 \). By applying our integration skills, we find the antiderivative of each term separately:

• For \( 2 \), the antiderivative is \( 2y \).
• For \( -\frac{y^2}{2} \), it becomes \( -\frac{y^3}{6} \).

The process involves finding the values at the upper and lower limits: \([ 2y - \frac{y^3}{6}]_{0}^{2}\).
Inserting the limits, we perform the subtraction:
\[\left( 2(2) - \frac{2^3}{6} \right) - \left( 2(0) - \frac{0^3}{6} \right)\]
Which results in \( 4 - \frac{8}{6} \), simplifying to \( \frac{8}{3} \). The outer integral finalizes the iterated process by combining all simplifying steps leading to a concrete numerical result.

Integral calculus often revolves around finding the **antiderivative**, and it is foundational in solving iterated integrals. An antiderivative essentially reverses differentiation; it is a function whose derivative matches the given one.
When given a simple function such as \( x \) or \( 2 \), finding the antiderivative is straightforward:

• For \( x \), the antiderivative is \( \frac{x^2}{2} \), meaning when you derive \( \frac{x^2}{2} \), you get back \( x \).
• For a constant like \( 2 \), integration becomes \( 2y \), aligning again with our derivative rules.

However, complexity can arise. In these scenarios, the polynomial expressions such as \( -\frac{y^2}{2} \) have antiderivatives like \( -\frac{y^3}{6} \).
Being adept at identifying and computing antiderivatives is crucial as it ensures successful resolution of iterated integrals. The final goal is to solve for a numerical value efficiently through precise integral evaluation.