Consider the function $f(x)$ (polynomial of degree two) described by \[ f(x) = ax^2 + bx + c. \] The derivative of this function is \[ f'(x) \equiv \frac{d}{dx}[f(x)] = \frac{d}{dx}\left[ ax^2 + bx + c \right] = \frac{d}{dx}\!\left[ ax^2\right] + \frac{d}{dx}\!\left[ bx \right] + \frac{d}{dx}\!\left[ c \right] = 2ax + b + 0, \] where we have used the linearity (derivative of a sum is equal to the sum of the derivatives of the individual terms) as well as the derivative formula $\frac{d}{dx}[x^n] = nx^{n-1}$. Recall that the derivative $f'(x)$ contains all the information about the rate of change or the slope of $f(x)$.

The inverse operation of the derivative is the anti-derivative (or integral), and it consists of adding up all the contribution of $f(x)$ (the area under the curve): \[ F(x) \equiv \int f(x)\ dx = \int ax^2 + bx + c\ dx = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + d. \] How do we know that $F(x)$ is the anti-derivative of $f(x)$? Well we just have to check whether $\frac{d}{dx}F(x) = f(x)$. If this is the case, then $F(x)$ is the anti-derivative of $f(x)$. Let's check: \[ \frac{d}{dx}F(x) = \frac{d}{dx}\left[ \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + d \right] = ax^2 + bx + c + 0 = f(x). \]

Ok so let's recap. Given a the formula for a function $f(x)$, your mad calculus skills allow you to calculate its derivative $f'(x)$ or its anti-derivative $F(x)$. This is full of WIN, because in physics there will be a lot of functions that will be thrown at you and you will have to calculate their derivatives and integrals.

Before we make this chapter more concrete and talk about physical quantities, I want to make some important observations. First note that differentiation destroys absolute information. In $f(x)$ we have the constant $c$, but after taking the derivative of this constant term it disappears. This is expected since the initial value of $f$ is of no importance, as far as the slope is concerned. We say that $f'(x)$ contains all the information of change in $f(x)$ but not its initial condition $c$. This is even more important when we consider the anti-derivative. If derivatives and integrals are inverse operations, then integration must create information. What does this mean? Well it means that calculus cannot tell you the value of the constant $d$. We could add any constant value to $F(x)$ and it will still be the anti-derivative of $f(x)$ since the derivative of a constant is zero. \[ \frac{d}{dx}(F(x) + 2929) = \frac{d}{dx}\left[ \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + d + 2929 \right] = ax^2 + bx + c + 0 = f(x). \] In other words, every time you integrate you get one free variable, which you can set to any value. As far as calculus is concerned $d$ or $d+2929$ are equally good constant terms for the anti-derivatives of $f(x)$. Thus, the anti derivative of $f(x)$ consists of a whole family of functions that look like $F(x)$ with some constant $d$ added on. In a physical problem, you will often be told some extra information that will allow you to pick the right $d$.