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Definition of $e$

$\ln (e)= \int_\limits{1}^e\frac{1}{t}dt=1$

I found this in my calculus textbook, and I was wondering what does this exactly "mean". This is more of a research question which I could do, but where would I start. I understand that its an integral, and what it states; however, how was this derived in calculus? Where would be a good book to research to understand an ideal proof of the equation?

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  • $\begingroup$ "Derived" may not be the correct word to use. What you posted can be a "definition" of the natural logarithm (the antiderivative $F$ of $\frac{1}{t}$ such that $F(e) = 1$). $\endgroup$ – Mark Nov 10 '16 at 0:41
  • $\begingroup$ Because $f(x)=\int_1^{x} \frac{1}{t}\,dt$ is strictly increasing, continuous, and unbounded, and $f(1)=0$, you know that $f(e)=1$ for some exactly one value $e>1$. $\endgroup$ – Thomas Andrews Nov 10 '16 at 0:44
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    $\begingroup$ Unless you already have another definition of $e$ of $\ln$, this is not a "theorem", i.e. something to be proved, but rather a definition. As such the only interesting thing to be proven about it i that it is well-defined, i.e. that there is a single number with this property. $\endgroup$ – Ian Nov 10 '16 at 0:49
  • $\begingroup$ @Ian So basically its something really basic that has no magical complex explanation. Because I understand it I just thought there would be a magical complicated way to show the why its a definition. $\endgroup$ – EnlightenedFunky Nov 10 '16 at 0:52
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    $\begingroup$ If this is how you're introducing $e$ for the first time then no there's nothing special there. What's striking is that this definition is equivalent to a bunch of other stuff; for example this $e$ is the same $e$ as $\lim_{n \to \infty} (1+1/n)^n$ from compound interest $\endgroup$ – Ian Nov 10 '16 at 0:58
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To say that $e$ is the number such that $\int_\limits{1}^e\frac1x\mathrm{d}x=1$ is to say that when you graph the curve of $y=\frac1x$ and shade in the area to the right of $x=1$, then $e$ is the boundary on the right such that the shaded area is exactly $1$. Take a look at the diagram below. The green curve is $y=\frac1x$ and $a$ is the area of the brown region.

enter image description here

If the boundary on the right were $x=3$, we'd have slightly too much area. If it were instead $2$, we'd have too little area. But $e=2.718\ldots$ is the point that's just right.

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