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The power rule for integration is one of the simplest and most commonly used tools in calculus. It allows you to find the antiderivative of a polynomial expression. The rule is: \[ \int x^n \, dx = \frac{1}{n+1}x^{n+1} + C \]where \(n eq -1\). The constant \(C\) represents the constant of integration, which appears because indefinite integrals usually have family of solutions. However, in definite integrals, you evaluate this expression across the limits of integration, which eliminates the constant. Let's look at how it applies here: when integrating \(x^2 y\) with respect to \(x\), you treat \(y\) as a constant. Using the power rule, the integral is:\[ \frac{1}{3}x^3 y \Big|_{0}^{2} \]Evaluating this expression at the limits \(x = 2\) and \(x = 0\) gives the result \(\frac{8}{3}y\). The key takeaway is to apply the power rule effectively to reduce the complexity of polynomial integrals.

Definite integrals compute the "net area" under a curve between two specified points along the x-axis or, in the case of iterated integrals, considering the whole surface below a given function part of a two-dimensional shape. Evaluating a definite integral requires finding an antiderivative of the integrand function and then calculating differences between its values at the upper and lower limits of integration.In the presented exercise, definite integration was applied twice. First, the integration over \(x\) from 0 to 2 was completed, resulting in some expression dependent on \(y\). Next, the outer integral involved integrating over \(y\) from 1 to 3, using the expression derived from the previous result. A straightforward integration of \(y\) resulted in:\[ \frac{8}{3} \int_{1}^{3} y \, dy \]When integrated using power rule, it leads to:\[ \frac{8}{3}\left[\frac{1}{2}y^2 \right]_{1}^{3} \]After evaluating this, the overall net area or volume found is \(\frac{32}{3}\). This illustrates how definite integrals provide concrete numerical answers representing physical quantities, like lengths, areas, or volumes.

Integration by parts is a powerful technique used mainly when dealing with integrals involving products of functions. The formula for integration by parts is derived from the product rule for differentiation and is given by:\[ \int u \, dv = uv - \int v \, du \]Where \(u\) and \(dv\) are chosen parts of the original function being integrated. Essentially, this method involves choosing one part of the product to differentiate and the other part to integrate, then solving the resulting expressions. Although integration by parts wasn't directly used in the original exercise, where the power rule was sufficient, knowing when and how to apply this technique is crucial, especially in more complex integrals. The idea is to "transfer" the difficulty from the original integral to a new one that might be easier to solve.When you encounter an integral that doesn't fit nicely into basic rules, considering integration by parts can offer a viable solution path. Understanding this principle broadens your toolkit for tackling a diverse range of integration challenges, including polynomials, exponentials, logarithms and trigonometric functions.