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In a given question I'm given the following function

$\int^{2x}_{x}\frac{1}{t}dt$

The question asks if the function is increasing or decreasing on the interval $(0,\infty)$

I've taken the integral of the function and got $\ln2$ but I'm not sure how to decipher whether the function is increasing or decreasing given this information.

Any tips?

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Well, the indefinite integral of $\frac{1}{t}$ is $\ln t+C$, so: $$ f(x)= \int_x^{2x}\frac{1}{t}dt= \ln(2x)-\ln(x)= \ln\left(\frac{2x}{x}\right)= \ln(2) $$ Therefore, $f$ is constant so it is both decreasing and increasing.

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  • $\begingroup$ does this mean that on the interval $(0,\infty)$ the function is increasing? what do you mean when you say it is both increasing and decreasing when the integral is a constant? $\endgroup$ – esc1234 Apr 20 at 13:56
  • $\begingroup$ @esc1234 $f$ is constant in that interval. If you examine the definition, you’ll see that a constant function is both increasing and decreasing. $\endgroup$ – Yuval Gat Apr 20 at 14:01
  • $\begingroup$ i marked your answer as the answer to my original question. The key part of the answer for me was that "$f$ is constant so it is both decreasing and increasing. But I am still not sure where this information is coming from. What definition are you referring to that states if the integral of a function is a constant it is both increasing and decreasing? thanks for your feedback btw $\endgroup$ – esc1234 Apr 20 at 20:58
  • $\begingroup$ @esc1234 you need to forget about the integral. $f$ is defined by an integral, but after evaluating we saw that it is identically equal to $\ln 2$. The definition of an increasing function: if $x<y$ in the domain, then $f(x)\le f(y)$. Here, $f\equiv\ln 2$, so for all $x, y\in (0, \infty)$ such that $x\le y$, we have $f(x)=\ln 2\le\ln 2=f(y)$ and hence $f$ is increasing there. Similarly, we obtain $f$ is decreasing there.. $\endgroup$ – Yuval Gat Apr 20 at 21:03
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$\int_{x}^{2x}{\frac{dt}{t}}=\log{2}$ therefore it is a constant function or both increasing and decreasing.

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Yes, the function is constant and it is always equal to $\log2$. So, it is both increasing and decreasing.

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