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While analyzing a certain function I was asked to integrate $$\int \left|f'\left(x\right)\right|f\left(x\right)dx$$ Since I don't know how to integrate an absolute value function, I noticed that it could be re-written as $$\int \sqrt{f'\left(x\right)^2}f\left(x\right)dx$$ and evaluated that integral, which gave me the correct answer. My question is, can you do that for every function? And how is $\sqrt{f'\left(x\right)^2}$ different than $f'\left(x\right)$? It's supposed to be just $$\left(f\left(x\right)^2\right)^{\frac{1}{2}} = f\left(x\right)$$ Isn't it?

libanobovich
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    $\sqrt{x^2}=|x|$ is true. $\sqrt{x^2}=x$ is, in the most general cases, false – FShrike Jun 06 '22 at 19:46
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    The general strategy for integrating absolute value functions is in evaluating separately on the positive and negative regions. – FShrike Jun 06 '22 at 19:47
  • Speaking to your last question, $(x^a)^b = x^{ab}$ is an identity when $b$ is an integer, but as soon as you consider more general exponents, you'll quickly find cases where that equality fails. This is required as soon as we define $x^a$ as a function of $x$. – Brian Moehring Jun 06 '22 at 20:17
  • @BrianMoehring: $x^a$ as a function of $x$, or $a^x$ as a function of $x$? – Brian Tung Jun 06 '22 at 21:56
  • @BrianTung For that equality, the important fact is we define $x^a$ as a function of $x$. For instance, setting $a=1/2$, we state that $x^{1/2}$ is a function of $x$. If I set $x=4$ in this function, I only get one value out. Depending on your decision, it could be $2$ or it could be $-2$, but once I make that decision, it's fixed. I can't have it equal $-2$ in one line and $2$ in the next because then I'm talking about two different functions. – Brian Moehring Jun 06 '22 at 22:21
  • More abstractly, call $x \mapsto x^a$ the function $\varphi_a$. Then $(x^a)^b = x^{ab}$ is the statement $\varphi_b \circ \varphi_a = \varphi_{ab}$. Once we define these three functions, we have equality or not, and some functions, like $\varphi_{1/2}$, have more than one candidate. – Brian Moehring Jun 06 '22 at 22:24
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    Your last question leads me to suspect you took an unwarranted shortcut in evaluating $\int \sqrt{f'\left(x\right)^2}f'\left(x\right)dx.$ Perhaps it turns out you're taking the square of something that is already guaranteed to be positive. What particular function is $f'$ in this integral? – David K Jun 06 '22 at 23:10
  • @BrianMoehring: I think I just misunderstood what you were saying, thinking that you were claiming that defining $x^a$ (as a function of $x$) was a problem for any $a$. – Brian Tung Jun 07 '22 at 03:15

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