Antiderivatives and Initial Values

An accelerometer can describe how velocity changes. Recovering velocity also requires an initial velocity; recovering position requires another initial condition. This is a general feature of reversing differentiation: a derivative loses information about additive constants.

An antiderivative of $f$ on an interval is a differentiable function $F$ satisfying $F'=f$ there. If $F$ is one antiderivative, every other one on the same interval has the form $F+C$, by the mean value theorem. We write

$$\int f(x)\,dx=F(x)+C.$$

This indefinite integral denotes a family of functions, not yet the definite integral of accumulation introduced next. The symbol $dx$ records the variable of integration. Changing it without changing the function’s interpretation can change the problem.

Reverse rules and verify by differentiation

For $r\ne-1$, on an interval where the real powers and derivatives are defined,

$$\int x^r\,dx=\frac{x^{r+1}}{r+1}+C.$$

The exceptional exponent matters: division by $r+1$ fails at $r=-1$. Instead $\int(1/x)\,dx=\log|x|+C$ on an interval avoiding zero. The disconnected domain permits different constants on the positive and negative half-lines.

Other elementary pairs are $e^x$, whose antiderivative is $e^x$; $\cos x$, whose antiderivative is $\sin x$; and $\sin x$, whose antiderivative is $-\cos x$. Linearity allows termwise integration of finite linear combinations. Products do not integrate by multiplying antiderivatives: $(\int f)(\int g)$ differentiates with a product-rule extra term.

For example,

$$\int(6x^2-4x+3)\,dx=2x^3-2x^2+3x+C.$$

Differentiate the answer to check every term. A symbolic answer that differentiates to the wrong expression is wrong regardless of how plausible its form looks.

Initial-value information

Suppose $F'(x)=3x^2-2$ and $F(1)=5$. Antidifferentiation gives $F=x^3-2x+C$. At $1$ this is $C-1$, so $C=6$. Thus $F=x^3-2x+6$. Without the initial value, there would be infinitely many vertical translates satisfying the derivative equation.

For a constant acceleration $a(t)=-9.8$ metres per second squared, an initial upward velocity $v(0)=19.6$ gives $v(t)=19.6-9.8t$. If $s(0)=1.5$ metres, the position is $s(t)=1.5+19.6t-4.9t^2$, over the time interval where the free-flight model applies. At the highest point $v=0$, so $t=2$ seconds and $s=21.1$ metres. The velocity initial condition and position initial condition determine two separate constants.

Direction changes and distance

For $v(t)=2t-2$ on $[0,3]$, a position antiderivative is $s(t)=t^2-2t+C$. Displacement is $s(3)-s(0)=3$ units of length. Velocity is negative on $[0,1)$ and positive on $(1,3]$, so the object first moves backward and then forward. Its distance from time zero to one is $|s(1)-s(0)|=1$, and from one to three is $|s(3)-s(1)|=4$, giving total distance $5$.

Taking the absolute value of final displacement would give $3$ and miss the reversal. Integration of speed will supply a uniform way to handle such cases. First identify where the sign changes so that the absolute value can be treated correctly.

Exercises

1. Find all antiderivatives of $4x^3-2x+5$ on $\mathbb R$.
2. Find an antiderivative of $x^{-2}+x^{-1}$ on $(0,\infty)$.
3. Solve $F'=2x+3$ with $F(0)=-4$.
4. An object has acceleration $6t$, initial velocity $-2$, and initial position $5$. Find velocity and position.
5. For $v(t)=t-2$ on $[0,5]$, find displacement and total distance.
6. Explain why one constant need not describe antiderivatives across a disconnected domain.

Worked solutions

1. $x^4-x^2+5x+C$.
2. $-1/x+\log x+C$; differentiate to verify both terms.
3. $F=x^2+3x-4$.
4. $v=3t^2-2$ and $s=t^3-2t+5$.
5. A position function is $t^2/2-2t$. Displacement is $2.5$; backward distance before $2$ is $2$, forward distance after is $4.5$, giving total distance $6.5$.
6. MVT connects two inputs through an interval in the domain. It cannot cross a missing point, so each component interval may have a separate constant.