Integral Equation (or I think so) Calculus I problem

[SqueeSpleen]

Homework Statement

Find a continuous funciton $f$ such that

$$
f(x) = 1+ \dfrac{1}{x} \int_{1}^{x} f(t)dt
$$
I think I solved it but I would like to see if it's right.
Well, first of all, by the fundamental theorem of calculus I know that
$$
\left( \int_{1}^{x} f(t)dt \right) ' = f(x)
$$

What I did was differentiate the original equation, to get a differential equation
$$
f'(x) = \left( -\dfrac{1}{x^{2}} \right) \int_{1}^{x} f(t)dt + \dfrac{1}{x} \left( f(x) \right)
$$
$$
f'(x) = -\dfrac{1}{x} \left( \dfrac{1}{x} \int_{1}^{x} f(t)dt \right) + \dfrac{1}{x} \left( f(x) \right)
$$

Well, by our first equation we have
$$
\dfrac{1}{x} \int_{1}^{x} f(t)dt = f(x)-1
$$
So using that we have
$$
f'(x) = -\dfrac{1}{x} \left( f(x)-1 \right) + \dfrac{1}{x} \left( f(x) \right)
$$
$$
f'(x) = -\dfrac{1}{x} f(x)+\dfrac{1}{x} + \dfrac{1}{x} \left( f(x) \right) = \dfrac{1}{x}
$$
So $f'(x)=1/x$. Before I started this thread, I verified it a good amount of times and I always reached $f'(x)=1$, so I guess I found at least one mistake in what I was doing. Wrote it to a forum really helps me to put my ideas together.
We integrate and we have
$$
f(x) = ln(x)+c
$$
Now when this is true.
$$
f(x) = 1+ \dfrac{1}{x} \int_{1}^{x} f(t)dt
= 1 + \dfrac{1}{x} \left (xln(x)-x+c \right)
= 1 + ln(x)-1+c
=1+ln(x)-1+\frac{c}{x}
$$
So $c=0$. Other way to check this is to notice that $f(1)=1$, so $c=0$ because the integral is zero when $x=1$.

Is this right? Another thing, I had problems with the TeX in the forum, half of it doesn't want to compile and I really don't know why. I don't know how to use the old [tex] [\tex], only how to use double dollar symbol to make an entire row of latex.

<Moderator's note: technical edit on LaTeX code. fresh_42. Please correct eventual mistakes.>

 

[Charles Link]

One mistake I found is that for $f(x)=ln(x)+c$ , $c=1$ (not zero). (Your Latex is hard to read. I think you missed a Latex parameter somewhere.) $\\$ (And I checked this solution by integrating the right side. It checked.)

 

[SqueeSpleen]

Sorry, initially it didn't even compile and after trying for a while and being unable to find what was wrong I posted it anyway. Moderator corrected the mistakes but I still don't know which they were. Yes, it has a lot of sense because $ln(1)=0$, so $c$ must be $1$, but I don't know where I commited the mistake that made $c=0$. I'll re-read this.

 

[Charles Link]

Sorry, initially it didn't even compile and after trying for a while and being unable to find what was wrong I posted it anyway. Moderator corrected the mistakes but I still don't know which they were. Yes, it has a lot of sense because $ln(1)=0$, so $c$ must be $1$, but I don't know where I commited the mistake that made $c=0$. I'll re-read this.

$f(1)=1$ so that $ln(1)+c=1$. But $ln(1)=0$ so that $c=1$.

 

[SqueeSpleen]

Yes but I'm having the following problem. I arrived to:
$$
f(x) = ln(x)+\frac{c}{x}
$$
Which implies that $c=0$, which is not true (for the thingsd you're saying). But... I'm not really sure where did I make the mistake.
Edit: Found the mistake, I have to integrate c too, I put it outside the integral as it was the case of an indefinite integral.

 

[Charles Link]

Yes but I'm having the following problem. I arrived to:
$$
f(x) = ln(x)+\frac{c}{x}
$$
Which implies that $c=0$, which is not true (for the thingsd you're saying). But... I'm not really sure where did I make the mistake.

That equation that you just edited needs a $cx$ instead of $c$ which simply makes both sides equal.

 

[SqueeSpleen]

$$
f(x) = 1+ \dfrac{1}{x} \int_{1}^{x} f(t)dt
= 1 + \dfrac{1}{x} \left (xln(x)-x+cx \right)
= 1 + ln(x)-1+c
=1+ln(x)-1+c
$$
Well, this achieves nothing but verification, so this equation was redudant (but at least it's not contradictory as it was with the mistake).
Now, $c=1$ because the of what you said, that's caused because the integral nullifies when $x=1$ as it's an integarl over a point which is always zero (at least when it's a riemann integral as is the case and always is in Calculus I).

Edit: Thank you very much!

 

[hunt_mat]

Multiply the initial equation through by $x$ to obtain
$$xf(x)=x+\int_{1}^{x}f(t)dt$$

Then simply differentiate using the fundamental theorem of calculus and the product rule to obtain:

$$f'(x)=\frac{1}{x}$$

 

[Ray Vickson]

$$
f(x) = 1+ \dfrac{1}{x} \int_{1}^{x} f(t)dt
= 1 + \dfrac{1}{x} \left (xln(x)-x+cx \right)
= 1 + ln(x)-1+c
=1+ln(x)-1+c
$$
Well, this achieves nothing but verification, so this equation was redudant (but at least it's not contradictory as it was with the mistake).
Now, $c=1$ because the of what you said, that's caused because the integral nullifies when $x=1$ as it's an integarl over a point which is always zero (at least when it's a riemann integral as is the case and always is in Calculus I).

Edit: Thank you very much!

It would have been much faster and much more straightforward to look at $F(x) = \int_1^x f(t)\, dt$, and note that it satisfies the DE
$$\frac{dF(x)}{dx} = 1 + \frac{1}{x} F(x), \; \; F(1) = 0.$$
The solution is easily obtained as $F(x) = x \ln x$, and from that you get $f(x)$ as $f(x) = F^{\prime}(x)$.

Note: please do not write $ln(x)$ or $ln x$; it looks ugly. The correct LaTeX form is $\ln(x)$ or $\ln x$, which is obtained by writing "\ln" instead of "ln". The same applies to other familiar functions like log, sin, cos, tan arcsin, arccos, arctan, sinh, cosh, tanh, lim, sup, inf, max, min, exp, gcd, etc. TeX/LaTeX is designed to work properly when these are preceded by a "\".

 

[SqueeSpleen]

Yeah, but I was trying to let this elementary. As they hadn't studied DE formally in this course, I didn't want to mention that's a DE.
But in the end what I'm doing is variation of parameters...
But yeah, your solution and huntman's one are way more straightforward, and even if I wanted to keep it elementary I could have done as you say without neccesarily introducting the concept of differential equation.

Thanks for the answers, if someone ask me again about those problems I will use your solutions as they're way easier to follow than my mess.