Prove p must be the zero function. $$p(x) = \int_0^x p(t) \,dt$$
I began by differentiating both sides:
$p'(x) = p(0) - p(x)$
From here, I am not sure how to go further to prove that it's a zero function.
 A: As noted by @Calvin_Khor, you differentiated incorrectly. Differentiating both sides gives you
$$p(x)=\int_0^x p(t)dt$$
$$p'(x)=p(x)$$
Now, the solution to this ODE can be solved by separating the equation
$$\frac{dp}{dx}=p$$
$$\frac{dp}{p}=dx$$
$$\int \frac{1}{p}dp=\int dx$$
$$\ln(p)=x+c$$
$$p(x)=Ae^x$$
Then plugging this back into the original equation gives
$$Ae^x=\int_0^x Ae^tdt=Ae^x-Ae^0=Ae^x-A$$
$$0=-A$$
$$A=0$$
We conclude that $p(x)=0$.
A: Suppose $p(x)$ is continuous there. Differentiate both sides with respect to $x$, and use fundamental theorem of calculus, which gives us $$\frac{dp(x)}{dx}=\frac{d}{dx}\int_0^{x}p(t)dt=p(x)-p(0).$$Here$\frac{dp(x)}{dx}=p(x)-p(0)$ is an ordinary differential equation. Firstly suppose that $p(x)-p(0)\neq 0$, hence we can solve it by separating $p$ and $x$, to be more specified$$\frac{dp(x)}{p(x)-p(0)}=dx$$ Integrate both sides and we obtain $$\text{log}|p(x)-p(0)|=x$$This gives $|p(x)-p(0)|=\mathrm{e}^x$. But when $x=0$ this implies $0=\mathrm{e}^0=1$, which provides a contradiction.
Now suppose that $p(x)\equiv p(0)$, which is to say, $p(x)$ is a constant number over $\mathbb{R}$. We also use $x=0$ to check, yet to find that $$p(0)=\int_0^0p(t)dt=0,$$ hence $p(x)\equiv 0$.
