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## 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.>

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