$f:\mathbb R\rightarrow \mathbb R$ 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 $f(1)=?$  
Please help. Thanks in advance for your time.
 A: Hint:  First, what is $f(0)$?  Second, consider $f'(x)$.  You get a differential equation that way.
A: 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 $f(1)$.
A: 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
$$f(x) = e^{5 x}$$.
A: 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.
