91Ó°ÊÓ

Before diving into the actual calculation of a double integral, it's crucial to comprehend the region of integration. In our original exercise, the given double integral \[ \int_{0}^{1} \int_{x^{2}}^{x}(x+y) \ dy \ dx \] presents a region defined in the x-y plane. This region is bounded by:

• The vertical line at \(x = 0\)
• The vertical line at \(x = 1\)
• The curve \(y = x^2\)
• The line \(y = x\)

This construction outlines a curvy triangular area where each x-value within the interval \([0,1]\) corresponds to a y-range from \(x^2\) to \(x\). Grasping the geometric shape of the integration area helps us visualize the problem and guide our integration process.

Iterated integration is the approach of evaluating a double integral by simplifying it into a successive series of single integrals. In our exercise, this involved two steps. First, with respect to y:\[\int_{x^2}^{x} (x+y) \, dy \]For this, apply the continuous integration rule by treating x as a constant:

• Integrate \(x\) with respect to \(y\) to get \(xy\)
• Integrate \(y\) with respect to \(y\) to get \(\frac{y^2}{2}\)

Then, plug in the boundaries: \[ (xy + \frac{y^2}{2})\vert_{x^2}^{x} \]Evaluate the limits, simplify, and proceed to integrate the simplified function:\[\int_{0}^{1} (\frac{3x^2}{2} - x^3 - \frac{x^4}{2}) \, dx\]Iterated integration allows complex problems to be broken down into simpler, manageable parts, making it a powerful tool in calculus.

The fundamental theorem of calculus links the concept of differentiation with integration and is a key player in solving double integrals efficiently. After setting up the iterated integrals, the theorem comes into play when evaluating the definite integrals after integration. In our case:\[ (xy + \frac{y^2}{2})\vert_{x^2}^{x} = (x(x) + \frac{x^2}{2}) - (x(x^2) + \frac{(x^2)^2}{2})\]By applying the fundamental theorem:

• We first derive an antiderivative for the expression.
• Then evaluate this antiderivative at the bounds \(x^2\) and \(x\).
• Subsequently subtract the results to complete the integral.

This powerful theorem not only simplifies the process of integration but also underlines the deep connection between differentiation and integration. It leads to effectively concluding our original exercise to get the neat fraction \(\frac{3}{20}\). This guides us seamlessly from complex problem setups to practical solving techniques.