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Chapter 6: Problem 31

Have your CAS evaluate \(\int(\sqrt{1-x}+\sqrt{x-1}) d x .\) If you get an answer, explain why it's wrong.

Short Answer

Expert verified
The given integral is undefined for all real numbers because there is no real value of x that falls in the real range for both functions simultaneously. If a CAS gives an answer, it would be incorrect as it doesn鈥檛 consider the exclusive domains of the two functions under the integral.

Step by step solution

01

Identify the Domains of the Functions

Firstly, observe the two functions under the integral separately, which are \(\sqrt{1-x}\) and \(\sqrt{x-1}\). The real domain for the function \(f(x) = \sqrt{1-x}\) ranges from \(-\infty\) to \(1\), inclusive, and the real domain for the function \(g(x) = \sqrt{x-1}\) ranges from \(1\) to \(\infty\), inclusive. Clearly there is no overlap between the two domains thus, no real number substitution into x will make both functions real simultaneously.
02

Attempt to Evaluate the Integral

Though none of the x-values fall within the real range of both functions, we still evaluate the integral of each function within their respective domains. Thus, the original integral decomposes to the sum of separate integrals as follows: \(\int(\sqrt{1-x}+\sqrt{x-1}) dx = \int \sqrt{1-x} dx + \int \sqrt{x-1} dx\). However, remember that since there鈥檚 no overlapping domain between the two functions, no real number can satisfy both integrands simultaneously thus the sum is not meaningful.
03

Conclusion

In this case, if we obtain an answer from a CAS, it would be incorrect because it doesn鈥檛 consider the mutual exclusivity of the two functions within the integral. In other words, the integral is not possible or undefined for all real numbers as there is no real value of x that falls in the real range for both functions simultaneously.

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