Antiderivatives and Initial Values

Reversing a derivative

An antiderivative $F$ of $f$ on an interval satisfies $F'=f$ . If one exists, adding a constant gives another because a constant differentiates to zero. Conversely, any two antiderivatives on the same interval differ by a constant, by the mean value theorem.

The notation $\int f(x)\,dx=F(x)+C$ describes this family. It is not yet a numerical accumulated change over an interval. On disconnected domain intervals, independent constants may be needed.

Visual guide

VISUAL GUIDEAntiderivatives differ by a vertical shift

x² − 2
x²
x² + 2

The curves x² − 2, x² and x² + 2 all have derivative 2x. Their slopes agree at each input even though their heights differ. The integration constant selects one member of this family.

Build a basic table

Reverse familiar derivatives:

\int x^p\,dx=\frac{x^{p+1}}{p+1}+C\quad(p\ne-1),

\int\frac{dx}{x}=\ln|x|+C.

The power formula applies on an interval where the real powers are defined and differentiable as needed. Also $\int e^x dx=e^x+C$ , $\int\cos x dx=\sin x+C$ , and $\int\sin x dx=-\cos x+C$ . Integration is linear: constants factor out and sums can be integrated term by term.

Worked example: verify the family

For $f(x)=6x^2-4x+3$ , integrate each term to obtain $F(x)=2x^3-2x^2+3x+C$ . Differentiating this result returns the original function, providing a direct check. If $F(1)=7$ , then $3+C=7$ , so $C=4$ .

The initial condition determines the constant, not the variable. Substituting $x=1$ into the rate before integrating would replace a varying rate by one instantaneous value and solve the wrong problem.

Worked example: motion with two conditions

Suppose acceleration is $a(t)=6t$ m/s², initial velocity is $v(0)=2$ m/s and initial position is $s(0)=5$ m. Integrating once gives $v(t)=3t^2+C_1$ , hence $C_1=2$ . Integrating again gives $s(t)=t^3+2t+C_2$ , hence $C_2=5$ .

Both constants matter because two differentiations removed two pieces of information. Checking $s''=6t$ , $s'(0)=2$ and $s(0)=5$ verifies the whole solution rather than only its shape.

Limits of a formula table

Not every elementary integrand has an elementary antiderivative. This does not prevent defining an accumulation integral or evaluating it numerically. More complicated compositions require substitution; products and rational expressions often need the dedicated techniques in Calculus II.

A common trap is applying the power formula at $p=-1$ , which would divide by zero. Another is forgetting the chain factor when checking a proposed answer. The derivative of $\sin(2x)$ is $2\cos(2x)$ , so an antiderivative of $\cos(2x)$ needs a factor of $1/2$ .

Practice

Integrate $4x^3-2$ .
Find $F$ if $F'=1/x$ on $(0,\infty)$ and $F(1)=3$ .
Find velocity if $v'=4$ and $v(0)=-2$ .

Show worked solutions

$x^4-2x+C$ .
$F=\ln x+C$ on the positive interval. The condition gives $C=3$ .
$v(t)=4t-2$ . Its derivative and initial value satisfy both requirements.