Question: I got this question:

Let $f$ be the integrable function defined on the interval $[0,2]$ by the rule:

$f(x)=
\begin{cases} 
4x^3 & \text{if $0 \leq x \leq 1$} \\
x^2+2 & \text{if $1<x \leq 2$} \\
\end{cases}$

Does there exist real numbers $c1$ and $c2$ such that the function $F$ defined below is an antiderivative for $f$ (i.e. $F'=f$)?

$F(x)=
\begin{cases} 
x^4 + c\_{1} & \text{if $0 \leq x \leq 1$} \\
\frac{x^3}{3}+2x+c\_{2} & \text{if $1<x \leq 2$} \\
\end{cases}$

I got stuck since the function $f$ got step/jump discontinuity at $x=1$.
And I don't know how to proceed from there.

Thanks.

Answer: 
I'm afraid that from a pre-calculus point of view, there is no consistent definition of the functional relationships that allows to discuss them as real functions (real valued in one real variable).

The big gap to be closed is the notion of an angle. In geometry, an angle is defined by a pair of rays from the same point. These most often occur at corners of triangles. All geometrically similar pairs of rays define the same angle. You can arrange such an equivalent pair of rays to have the first ray horizontal pointing to the right, then the second ray is uniquely characterized by a point on the unit circle (after fixing an origin and a unit length). The coordinates of that point then are the cosine and sine of that (equivalence class of) angle(s).

Then there are some angles that can be constructed and calculated geometrically by constructing regular n-gons and thus dividing the unit circle into n equal parts. n=3,4,6,8,12 are easy, n=5 is possible without too much effort. By bisection one can enclose any other angle in fractions of the full circle, starting from these known fractions, but this goes close to limit arguments, which already is calculus.

One can identify points on the unit circle with rotation matrices or complex numbers, which simplifies the angle arithmetic. Doubling an angle is then squaring the matrix or complex number, bisecting an angle corresponds to a square root. This gives functional equations that in their components are the trigonometric identities. Demanding continuity and differentiability at the zero angle will also give the trigonometric functions as unique solutions. However, this uniqueness is again enforced by calculus arguments.
If you want to verify this intuitively for yourself, then this answer could prove to be helpful. But remember, *this is not a proof*.

First try to visualize what $\sin\theta$ looks like. If you need some help, this is what it looks like.

[![Alt text](https://i.stack.imgur.com/vLA0e.gif)](https://i.stack.imgur.com/vLA0e.gif)

Now remember that if $\sin\theta$ is a function, then for a certain $\theta$, there can be only *one* value of $\sin\theta$.

Now pick a random point on the $x-$axis. This is going to be your value for $\theta$. Now a draw a line perpendicular to the $x-$axis through that point. Now *is it possible for that line to intersect the sine curve at more than one point?* Try it out for different values of $x$. No, it is completely impossible for that line to intersect that curve at more than one point.

This means that for a certain $\theta$, there is only one associated value of $\sin\theta$.

$\sin\theta$ and $\cos\theta$ are pretty much the same. Notice that $\cos\theta=\sin(\theta +\frac{\pi}{2})$. So if you move the sine curve by $\frac{\pi}{2}$ to the right, you're going to get the curve for $\cos\theta$. And $\sin\theta$ and $\cos\theta$ *look* exactly the same. So you can use the same argument.

Finally, I repeat that this is not a proof. But I hope this helps you see it intuitively.

The figure is taken from [here](http://jwilson.coe.uga.edu/emat6680/dunbar/Assignment1/sine_curves_KD.html).