Given $f’(x) = x^2$, derive $f(x) = \frac{x^3}{3} + C$ without integration Suppose that $f$$:$$\mathbb{R}\to\mathbb{R}$ is differentiable at every point and that 
$$f’(x) = x^2$$
for all $x$. Prove that 
$$f(x) = \frac{x^3}{3} + C$$
where $C$ is a constant. 
This has to be done without integrating, I have only been taught differential calculus and this question only assumes knowledge of that.
I tried applying mean value theorem and taylor’s approximation but could not come up with the proof. Can someone please provide the solution?
 A: Let $g:\mathbb R\to\mathbb R$ be differentiable such that $g'(x) = 0$ for all $x\in\mathbb R$. Then the mean value theorem says that for all $x,y\in\mathbb R$ there exists some $\xi$ between them such that $g(x)-g(y) = g'(\xi)(x-y) = 0$. Hence, $g(x)=g(y)$ for all $x,y\in\mathbb R$ and it follows that $g$ is constant.
Now, consider the function $\phi(x) = \frac 13 x^3$. Its derivative is $\phi'(x) = x^2$. But also your function $f : \mathbb R\to\mathbb R$ has that derivative. Consider $g := f-\phi$. We have $g'(x) = f'(x)-\phi'(x) = x^2-x^2 = 0$ for all $x$, so $g$ is a constant, $g(x) = c$, and thus $f(x) = \phi(x) + g(x) = \frac 13x^3 + c$.
A: Let $g(x) = f(x) - \frac{x^3}{3}$.  What can you say about the derivative of $g(x)$?
A: Here is to use the Mean-Value-Theorem (MVT) for the derivation. Let $\epsilon$ be small positive variable and observe that,
$$x^2+\frac13\epsilon^2= \frac{6\epsilon x^2+2\epsilon^3}{6\epsilon}
=\frac{(x+\epsilon)^3 - (x-\epsilon)^3}{3(2\epsilon)}
=\frac{g(x+\epsilon)-g(x-\epsilon)}{(x+\epsilon)-(x-\epsilon)}\tag{1}$$
where 
$$g(x) = \frac13 x^3$$
According to MVT, the equation (1) can be written as, 
$$\frac{g(x+\epsilon)-g(x-\epsilon)}{(x+\epsilon)-(x-\epsilon)}=g'(a)=x^2+\frac13\epsilon^2$$
where $x-\epsilon < a < x+\epsilon$. Now, let $\epsilon \rightarrow 0$ to obtain,
$$g'(x)=x^2=f'(x)$$
Therefore,
$$f(x) = g(x) +C= \frac13 x^3+C$$
