# Proving area under the integrals.

I have a question that I have been trying to solve that I am curious about. If you have a continuous function $f(x) = \frac1x$.

How would you prove that $$\int_1^af(x)\,dx+\int_1^bf(x)\,dx=\int_1^{ab}f(x)\,dx$$ assuming that $a > 1$ and $b > a$?

Here is a proof that does not use logarithms or substitution, but rather the fundamental theorem of calculus and the chain rule.

On the one hand, $$\frac{d}{da}\left(\int_1^af(x)\,dx+\int_1^bf(x)\,dx\right)=\frac1{a}$$

On the other hand, $$\frac{d}{da}\int_1^{ab}f(x)\,dx=\frac{1}{ab}\cdot b=\frac{1}{a}$$

So the two functions of $a$ have the same derivative, and therefore differ by a constant with respect to $a$ (that may depend on $b$). That is, $$\int_1^af(x)\,dx+\int_1^bf(x)\,dx=\int_1^{ab}f(x)\,dx+C_1(b)$$ where $C_1(b)$ is constant with respect to $a$.

The same argument using $\frac{d}{db}$ shows $$\int_1^af(x)\,dx+\int_1^bf(x)\,dx=\int_1^{ab}f(x)\,dx+C_2(a)$$ from which we can deduce $C_1=C_2$ is a constant function with respect to both $a$ and $b$. Let $a=b=1$, and we conclude this constant function is the zero function.

To see why $\ln ab = \ln a + \ln b$, observe that: \begin{align*} \int_1^{ab} \frac{1}{x}dx &= \int_1^{a} \frac{1}{x}dx + \int_a^{ab} \frac{1}{x}dx \end{align*} Now for the last integral, make the substitution $u = x/a$ so that $du = dx/a$, giving us: $$\int_a^{ab} \frac{1}{x}dx = \int_1^{b} \frac{1}{au}(a \, du) = \int_1^{b} \frac{1}{u}du = \int_1^{b} \frac{1}{x}dx$$

The easiest method is just to solve the integrals and see if $LHS=RHS$.

$$\int_1^af(x)\,dx+\int_1^bf(x)\,dx=\int_1^{ab}f(x)\,dx$$

Proof:

LHS:

$$\int_1^a\frac1x\,dx+\int_1^b\frac1x\,dx=\ln a-\ln1+\ln b-\ln 1=\ln a+\ln b$$

RHS:

$$\int_1^{ab}\frac1x\,dx=\ln ab-\ln1=\ln ab=\ln a +\ln b$$

Therefore $$\int_1^a\frac1x\,dx+\int_1^b\frac1x\,dx=\int_1^{ab}\frac1x\,dx$$

To prove only using geometry, consider a transformation of the area represented by $$\int_1^b\frac1x\,dx$$ where you scale that area away from the $y$-axis by a factor of $a$, and simultaneously scale it towards the $x$ axis by a factor of $\frac1a$. The net area will not be different, but it will be in a different place. You will now have $$\int_a^{ab}\frac1x\,dx$$ and now it is clear why this can be added to $\int_1^a\frac1x\,dx$ to get $\int_1^{ab}\frac1x\,dx$.

(Since the question states that we can assume $a,b>1$, I wonder if something like this was intended. The various other proofs work fine for $a,b>0$. But this one is not quite so clean for $a,b$ in $(0,1)$, since negative areas would come into play.)

It is the same like you try to prove that

$Ln(a) + Ln(b) = Ln(ab)$

.............

Assume :
$x = Ln(a)$ and $y = Ln(b)$

which means :

$a = e^x$ and $b = e^y$

Now multiply :

$$ab = e^x * e^y$$

$$ab= e^{x+y}$$

Now introduce the logarithm on both sides :

$$Ln(ab) = Ln(e^{x+y})$$

$$Ln(ab) = x + y$$

$$Ln(ab) = Ln(a) + Ln(b)$$

We will assume certain facts about integrals that may not have been proved yet in your course. We have $$\int_1^{ab}\frac{1}{x}\,dx=\int_1^a \frac{1}{x}\,dx+\int_a^{ab}\frac{1}{x}\,dx.\tag{1}$$ For the second integral, make the substitution $x=at$. Then $dx=a\,dt$, and $$\int_a^{ab}\frac{1}{x}\,dx=\int_1^b \frac{1}{at}a\,dt=\int_1^b \frac{1}{x}\,dx.$$

First integrate the function $f(x)$.

The result is a logarithmic function.

If you solve using this you will get

$$\ln a + \ln b = \ln (ab)$$