# $f:\mathbb R\rightarrow \mathbb R$ be a continuous function satisfying $f(x)=5\int_{0}^{x}f(t)dt+1$ for all $x \in \mathbb R$

I was thinking about the following problem:

Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function satisfying $f(x)=5\int_{0}^{x}f(t)dt+1$ for all $x \in \mathbb R$.Then $f(1)=?$

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$f(t)=5\int_{0}^{t}f(t)dt+1$ this is devoided of meaning. Do you understand why? –  Git Gud Jan 10 '13 at 17:12

From the point of view of the integral $\int_0^xf(t)\,dt$, $x$ is a constant that has nothing to do with $t$; $f(1)=\int_0^1f(t)\,dt$, and $\int_0^1f(1)\,d(1)$ is meaningless. Let’s start over with

$$f(x)=5\int_{0}^{x}f(t)dt+1\;.$$

Use the fundamental theorem of calculus and differentiate this with respect to $x$ to get

$$f\,'(x)=\frac{d}{dx}\int_0^xf(t)\,dt=5f(x)\;.$$

Now you know that $$\frac{f\,'(x)}{f(x)}=5\;,$$

where the lefthand side is the derivative of $\ln f(x)$ with respect to $x$. In other words,

$$\frac{d}{dx}\ln f(x)=5\;.$$

Take the antiderivative of each side to get

$$\ln f(x)=5x+C$$ and exponentiate: $f(x)=e^{5x+C}=ae^{5x}$, where $a=e^C$. Thus, $f(1)=ae^5$, and all that remains is to discover $a$. To do this, go back to the original definition of $f(x)$ and substitute $x=0$; what do you get for $f(0)$? Now substitute that into $f(0)=ae^{5\cdot0}=ae^0=a$ to discover what $a$ is, and you’ll have a specific numerical value for $f(1)$.

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Could we solve this problem by using Laplace transformation?? I am thinking about it. I see we have $f(0)=1$.Thanks, Brian. +1 for your nice approach. Please have a look at mine below.;-) –  B. S. Jan 10 '13 at 20:01

Hint: First, what is $f(0)$? Second, consider $f'(x)$. You get a differential equation that way.

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Is mine as you suggested? Thanks. –  B. S. Jan 10 '13 at 20:31
@BabakSorouh: Yes, I agree with your answer. I was hinting toward the approach that Brian M. Scott used, but the result is the same. –  Ross Millikan Jan 10 '13 at 20:48
Thanks Ross for the reply. +1 –  B. S. Jan 11 '13 at 12:57

No. Sorry, your answer is nonsense and demonstrates a weak grasp of calculus.

What you want to do is take a derivative with respect to x:

$$f'(x) = 5 f(x)$$

with $f(0) = 1$.

I'd tell you to solve this, but I think you may need to see a lot of steps. The solution is

$$f(x) = e^{5 x}$$.

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Thanks a lot sir.I got it.I am editing my post. –  user52976 Jan 10 '13 at 17:24
@user33640 Don't take the first comment to seriously! rlgordonna cannot derive that much about you by one mistake - possibly, that demonstrate a weak judge of people. –  AD. Jan 10 '13 at 17:26
@user33640: Yes, please, the comment is blunt, but not personal at all. You are likely courageous and resourceful for posting here. But in a forum like this where we take our subject seriously, we have a duty of harsh criticism of the subject matter. If your errors were caused by fast typing, etc., and not through weak knowledge, I apologize. –  Ron Gordon Jan 10 '13 at 17:33
I did not mind anything,sir.You are always welcome to criticize.There is so much to learn from all of you. Thanks again for your feedback. –  user52976 Jan 10 '13 at 17:37
You have $f(x)=5\int_0^xf(t)dt+1$. If you are familiar to Laplace Transformation, you have $$\mathcal{L}(f(x))=\mathcal{L}\left(5\int_0^xf(t)dt+1\right)=\mathcal{L}\left(5\int_0^xf(t)dt\right)+\mathcal{L}(1)=\\\ 5\mathcal{L}\left(\int_0^xf(t)dt\right)+\mathcal{L} (1)=5\mathcal{L}(1*f(x))+\mathcal{L}(1)=5\left(\frac{1}{s}\cdot\mathcal{L}(f(x))\right)+\frac{1}{s}$$ So $\mathcal{L}(f(x))=\frac{1}{s-5}$ that means $f(x)=\text{e}^{5x}$. Note that $1*f(x)$ means the convolution between $1$ and $f(x)$. Any way, you chose the simple and straightforward answer.
+1 $~~~~~~~~~~~~~~~~~~~~~~$ –  amWhy Feb 19 '13 at 0:15