An antiderivative is a whole family of functions, $F(x) + C$. One extra fact, such as a starting position, picks out a single member of the family. This is how Position, Velocity and Acceleration is recovered from acceleration.
Key idea
Why it works
Worked example: One condition
- $f(x) = 2x^3 - 4x + C$.
- $f(1) = 2 - 4 + C = -2 + C = 3$, so $C = 5$.
Answer: $f(x) = 2x^3 - 4x + 5$
Worked example: From gravity to height
- $v(t) = \int -32\,dt = -32t + C_1$, and $v(0) = 48$ gives $C_1 = 48$.
- $$s(t) = \int(-32t + 48)\,dt = -16t^2 + 48t + C_2$$, and $s(0) = 64$ gives $C_2 = 64$.
- Lands when $s = 0$: $-16t^2 + 48t + 64 = 0$, so $t^2 - 3t - 4 = (t - 4)(t + 1) = 0$, and $t = 4$.
Answer: $s(t) = -16t^2 + 48t + 64$; it lands after 4 seconds.
Worked example: Two conditions
- $f'(x) = 6x^2 + C_1$; $f'(0) = C_1 = 2$.
- $f(x) = 2x^3 + 2x + C_2$; $f(0) = C_2 = -1$.
Answer: $f(x) = 2x^3 + 2x - 1$
Problems like these are called initial value problems, and they are the simplest differential equations. A second-order problem needs two conditions because two antiderivatives bring two constants; in physics these are the starting position and the starting velocity. Every projectile under constant gravity follows $s(t) = -\frac{g}{2}t^2 + v_0t + s_0$, where the last two coefficients are exactly those initial values.
Common mistake