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I have the following function

$$g(x) = \int_0^{(\frac{x}{2})^2} \sin \sqrt{t}\ dt$$

Applying the fundamental theorem of calculus one would get

$$g'(x) = \frac{x}{2} \sin \sqrt{(\frac{x}{2}) ^2}$$

My question is related with the square root within the $\sin$ function. It would seem to me that the square root would cancel with the square and we would get

$$g'(x) = \frac{x}{2} \sin (\frac{x}{2})$$

However I was told that this was not quite right, and that the correct answer is

$$g'(x) = \frac{x}{2} \sin (\frac{|x|}{2})$$

I assume it has something to do with allowing the function to be defined over the whole set of the real numbers, but I am not being able to grasp why. Could someone please clarify?

Thanks in advance.

user1790813
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2 Answers2

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The integral and sine function are not important here. The point is that the square root sign is defined as the positive root, so $\sqrt {a^2}=|a|$. If we try it with $a=-2$, we have $a^2=4, \sqrt 4=2$ This is covered in a number of other questions on the site. One of them is here

Community
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Ross Millikan
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A very common misconception is that $\sqrt{x^2}$ can be simplified to $x$. But in fact they are only equal to each other for positive values of $x$. If $x$ is negative, then $\sqrt{x^2}=-x$.

In general, if you square a number and then take the square root of the result, you end up with the absolute value of the number you began with: i.e. the correct identity is $\sqrt{x^2}=|x|$.

mweiss
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  • So if $x$ is positive, then there is no problem with leaving the modulus out of the equation. However, does it even make sense to talk about the derivative of that integral when the $x$ is negative? That is what is really bothering me. – user1790813 Dec 29 '15 at 16:11
  • Sure. Notice that (for example) $g(-6)$ is defined to be an integral that has limits of integration from $0$ to $9$, an interval over which the integrand is defined. – mweiss Dec 29 '15 at 16:17