Question

State why it is difficult/impossible to integrate the iterated integral in the given order of integration. Change the order of integration and evaluate the new iterated integral. $$ \int_{0}^{\sqrt{\pi / 2}} \int_{x}^{\sqrt{\pi / 2}} \cos \left(y^{2}\right) d y d x $$

Short answer

The integral is \(\frac{1}{2}\) after changing the order of integration.

Step-by-step solution

Identify the Region of Integration

The given iterated integral suggests integration over the region defined by \(0 \leq x \leq \sqrt{\pi/2}\) and \(x \leq y \leq \sqrt{\pi/2}\). This region is a triangular area in the xy-plane.

Analyze the Integrand

The integrand is \(\cos(y^2)\), which is a non-elementary function, meaning that its integral does not have a simple closed-form representation. This complicates the integration in the given order.

Change the Order of Integration

To change the order, we need to consider the limits for \(y\) and \(x\):- \(y\) will range from \(0\) to \(\sqrt{\pi/2}\).- For a fixed \(y\), \(x\) will vary from 0 to \(y\), describing the same triangular region.Thus, the new integral becomes:\[\int_{0}^{\sqrt{\pi/2}} \int_{0}^{y} \cos(y^2) \, dx \, dy\]

Evaluate the New Iterated Integral

First, integrate the inner integral with respect to \(x\):\[\int_{0}^{y} \cos(y^2) \, dx = [x \cos(y^2)]_{0}^{y} = y \cos(y^2)\]Then, evaluate the outer integral:\[\int_{0}^{\sqrt{\pi/2}} y \cos(y^2) \, dy\]This integral can be solved efficiently via substitution:Let \(u = y^2\), then \(du = 2y \, dy\), or \(y \, dy = \frac{1}{2} du\).Now transform the limits: when \(y = 0\), \(u = 0\); when \(y = \sqrt{\pi/2}\), \(u = \pi/2\).Substituting these into the integral:\[\frac{1}{2} \int_{0}^{\pi/2} \cos(u) \, du\]This integral equals:\[\frac{1}{2} [\sin(u)]_{0}^{\pi/2} = \frac{1}{2} (1 - 0) = \frac{1}{2}\]

Conclusion

Re-evaluation by changing the order of integration allows for straightforward integration of \(\cos(y^2)\), which was initially difficult due to its limits with respect to \(x\). The reconstructed integral yields a result of \(\frac{1}{2}\).

Key concepts

Order of Integration

The term "order of integration" refers to the sequence in which integration is performed in the context of iterated integrals. In double integrals, this generally means deciding whether to first integrate with respect to one variable or the other.
When you're looking at an iterated integral, such as \[\int_{0}^{\sqrt{\pi / 2}} \int_{x}^{\sqrt{\pi / 2}} \cos \left(y^{2}\right) dy \, dx\] the order of integration is first in terms of \(y\), then \(x\).
Changing the order of integration can often make an integral easier to evaluate—especially when dealing with complex functions or regions. In this exercise, changing the order allowed the problematic integration of \(\cos(y^2)\) to become straightforward by handling the limits differently. The revamped integral turned into:\[\int_{0}^{\sqrt{\pi/2}} \int_{0}^{y} \cos(y^2) \, dx \, dy\]Now \(x\) is integrated out first, a much simpler prospect because its limits became constants which makes the integration of non-elementary functions over these bounds easier.

Triangular Region in Integration

In a double integral, the region over which you integrate can take many forms, and sometimes specifying the region clearly can simplify the work. The original integral setup describes a triangular region.
This triangular region is specified in the xy-plane by the bounds:

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

Visualizing this region can be helpful. Imagine it as a shaded area under the curve \(y = \sqrt{\pi/2}\) extending from the origin straight up to \((x = \sqrt{\pi/2}, y = \sqrt{\pi/2})\) and forming a right-angled triangle with the coordinates.
Re-ordering integration swaps the limits so that we integrate along rectangles from \(y = 0\) to its maximum, and in each stripe from \(x = 0\) to \(x = y\). By doing so, we preserve the total area of the original region but in a form that's potentially easier to work with mathematically.

Non-elementary Functions

A non-elementary function is one that does not have a simple closed-form integral. That means you can't express its integral using elementary functions like polynomials, exponentials, and logarithms.
In this exercise, the integrand \(\cos(y^2)\) is a non-elementary function. This kind of function is common and often requires special techniques to integrate. When faced with non-elementary functions in iterated integrals, changing the order of integration can sometimes circumvent the difficulty
By re-arranging the integral, we managed to sidestep the tricky direct integration of \(\cos(y^2)\) with respect to \(y\). Instead, integrating first with respect to \(x\) gave us expressions that were more manageable and allowed us to complete the integral efficiently, exploiting the structure of the substitution method to find the solution.