I am unable to interpret the integral
$$\int {1\over x}{\rm d}x=\ln|x|+\mathrm{C}$$
Graphically as area under the curve of $1/x$ (as the definition of the integral). Can somebody please illustrate/explain.
I understand this fact that derivative of $\ln(x)$ is $1/x$ and we take absolute value to make the support extend to entire real line, $x \neq 0$.
Another reason is that, if we start with the functional equation for the logarithm, $f(xy) =f(x)+f(y) $, we end up with the integral of $\frac1{x}$.
Here's how it goes:
We have $f(xy) =f(x)+f(y) $, and we want to find $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} $.
In what follows, I assume that $f(x)$ is as smooth as needed. In particular, I assume that $f'(x)$ exists for $x > 0$.
First, putting $x=y=1$, we get $f(1) = f(1)+f(1)$, so that $f(1) = 0$.
Then, $f(x+h) =f(x(1+h/x)) =f(x)+f(1+h/x) $, so that $f(x+h)-f(x) =f(1+h/x) $. Therefore
$\begin{array}\\ \dfrac{f(x+h)-f(x)}{h} &=\dfrac{f(1+h/x)}{h}\\ &=\dfrac1{x}\dfrac{f(1+h/x)}{h/x}\\ &=\dfrac1{x}\dfrac{f(1+h/x)-f(1)}{h/x} \quad\text{(since }f(1) = 0)\\ \end{array} $.
Letting $h \to 0$ on both sides, we get $f'(x) =\dfrac{f'(1)}{x} $, or
$\begin{array}\\ f(x) &=f(x)-f(1)\\ &=\int_1^x f'(t)dt\\ &=\int_1^x \dfrac{f'(1)dt}{t}\\ &=f'(1)\int_1^x \dfrac{dt}{t}\\ \end{array} $
If we choose $f'(1) = 1 $, which is a natural choice, we get $f(x) = \int_1^x \dfrac{dt}{t} $, which is one of the reasons this is called the natural logarithm.
By choosing different values for $f'(1)$, we can get the other logs: base 10, base 2, and so on.
As if often the case in my answers, nothing original, and I've said this before, but if the question isn't being flagged as a duplicate, then my answer shouldn't either.
Let's define a function $f$ as $f(x) = \int_1^{x} \frac{1}{t} dt$, $x>0$.
It is easy to see show the following properties:
Property (1) The function $f$ is increasing.
To show this, assume $x_2>x_1>0$, and form the difference $f(x_2)-f(x_1)=\int_{x_1}^{x_2} \frac{dt}{t}$. Note that this difference is strictly positive since the integrand is also strictly positive. Thus, $f(x_2)>f(x_1)$ whenever $x_2>x_1$, which implies $f$ is strictly increasing.
Property (2) The function $f$ is differentiable with $f'(x)=\frac{1}{x}$.
This follows immediately from the fundamental theorem of calculus. Note that this implies automatically that $f$ is continuous for $x>0$.
Property (3) The limit $\lim_{x \to \infty} f(x)=+\infty$
We can easily verify that $\lim_{x \to \infty} f(x)=+\infty$ since for $n>1$
$$\int_1^{2^n} \frac{1}{t} dt=\int_{1}^{2} \frac{1}{t} dt+\int_{2}^{4} \frac{1}{t} dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{t} dt$$
$$\ge \int_{1}^{2} \frac{1}{2} dt+\int_{2}^{4} \frac{1}{4} dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{2^n} dt=\frac12 (2-1)+\frac14 (4-2) + \cdots + \frac{1}{2^{n}}(2^n-2^{n-1})=\frac{n}{2}$$
which tends to infinity as $n \to \infty$.
Property (4) The limit $\lim_{x \to 0} f(x)=-\infty$
Examine the integral
$$f(\frac{1}{n})=-\int_{\frac{1}{n}}^1 \frac{dt}{t}$$ and substitute $u=\frac{1}{t}$. Then, we have $$-\int_{\frac{1}{n}}^1 \frac{dt}{t}=-\int_1^n \frac{du}{u}$$
But from property (3), this integral tends to $+\infty$ and thus $f(\frac1{n})$ tends to $-\infty$ as $n \to 0$.
Property (5) The function $f$ satisfies $f(x^n) = nf(x)$.
We know by definition that $$f(x^n) = \int_1^{x^n} \frac{1}{t} dt$$
Then, by substituting $u=t^n$, with $du=nt^{n-1}dt$, and the limits going from $1$ to $x$ reveals $$\log (x^n) =\int_1^{x} \frac{1}{u^{\frac{1}{n}}} nu^{\frac{1-n}{n}}du=\int_1^{x} n\frac{1}udu=nf (x)$$
NOTE:
The other well-known properties for the log, such as of a product and quotient rules, follow from Property (5).
We can also show that the inverse function, call it $exp(x)$, for $f(x)$ has all of the properties of an exponential function.
And finally, if we define the number $e=exp(1)$ such that $f(e)=1$, then we show that
$$e=\lim_{x\to 0}\left(1+x\right)^{1/x}$$
since $f'(1)=1=f(e)$ and also
$$\begin{align} f'(1)&\equiv\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\\\\ &=\lim_{h\to 0}f\left((1+h)^{1/h}\right)\\\\ &=f(e) \end{align}$$
which by continuity of $f$ implies $e=\lim_{h\to 0}\left((1+h)^{1/h}\right)$ and we are done!
If we define $F(x)=\int_1^x f(t)dt$ and differentiat it we see that $F'(x)=f(x) $ also $F'(ax)=f(x)$ then differenc of $F(x)$ and $F(ax)$ must be constant. After some calculation we find that $F$ is a logarithm that called natural logarithm. So $Ln(x)$ is the area of under of curve $y=\frac{1}{x}$ from $1$ to $x$ which is positive for $x>1$ and negative for $x<1$ and obviously $0$ for $x=1$.
You ask to demonstrate the result as a sum of areas, so ... Consider approximating the integral with $n$ "left-handed" Riemann rectangles with congruent bases, observing $$\int_{y}^{xy}\frac{dt}{t} \;\approx\; \sum_{k=0}^{n-1}\frac{\frac{1}{n}(xy-y)}{y+\frac{k}{n}(xy-y)} = \sum_{k=0}^{n-1}\frac{\frac{1}{n}(x-1)}{1+\frac{k}{n}(x-1)} \approx \int_1^{x} \frac{dt}{t} $$
Of course, as $n\to\infty$, the "$\approx$"s become "$=$"s, so that $$\int_{y}^{xy}\frac{dt}{t} = \int_1^{x} \frac{dt}{t}$$
Writing $L(x) := \int_1^x dt/t$, this implies $$L(x) + L(y) = L(xy)$$ so that the functional equation for the logarithm is satisfied by $L$. Determining that $L$ must be, specifically, the natural logarithm is left as an exercise to the reader.
