2
$\begingroup$

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 $b > a$?

$\endgroup$
1

7 Answers 7

6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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: $$ \int_a^{ab} \frac{1}{x}dx = \int_1^{b} \frac{1}{au}(a \, du) = \int_1^{b} \frac{1}{u}du = \int_1^{b} \frac{1}{x}dx $$

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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 :

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

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

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

$\endgroup$
0
$\begingroup$

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 $$\int_a^{ab}\frac{1}{x}\,dx=\int_1^b \frac{1}{at}a\,dt=\int_1^b \frac{1}{x}\,dx.$$

$\endgroup$
0
$\begingroup$

First integrate the function $f(x)$.

The result is a logarithmic function.

If you solve using this you will get

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.