Antiderivatives and indefinite integrals
“…the process of differentiation may be reversed…”
What does a derivative leave out?
Suppose a moving object’s velocity is known at every time. Can we recover its position? The derivative tells us how position changes, but it does not tell us where the object started. Two objects can have the same velocity throughout a journey while remaining a fixed distance apart.
The same issue appears without a physical setting. Each of
\[x^2,\qquad x^2+3,\qquad x^2-5\]has derivative \(2x\). Reversing differentiation should recover all of these possibilities. The missing information is an additive constant.
Thus \(F(x)=x^2\) is an antiderivative of \(f(x)=2x\) on \(\mathbb R\). To check a proposed antiderivative, differentiate it and compare the result with \(f\) throughout the stated interval.
Why adding a constant gives every solution
Theorem
Antiderivatives differ by a constant on an interval
If \(F\) is an antiderivative of \(f\) on an open interval \(I\), then every antiderivative of \(f\) on \(I\) has the form
\[F(x)+C,\]where \(C\) is a real constant. Conversely, each such function is an antiderivative of \(f\) on \(I\).
Proof (optional)
Let \(G\) be another antiderivative and put \(H=G-F\). Then \(H^{\prime}=f-f=0\) throughout \(I\). For any \(a<b\) in \(I\), the entire segment \([a,b]\) lies in \(I\). The function \(H\) is continuous on this segment and differentiable inside it, so the mean value theorem gives a point \(c\in(a,b)\) with
\[\begin{aligned} H(b)-H(a)&=H^{\prime}(c)(b-a)\\ &=0. \end{aligned}\]Therefore every pair of values of \(H\) agrees: \(H=C\), and \(G=F+C\). Conversely, differentiating \(F+C\) gives \(f+0=f\).
On an interval, a derivative determines the shape of a graph but leaves its vertical position undetermined.
The investigation below pairs a family of antiderivatives with their common derivative. The pale curves show three fixed members of the family; the darker curve is the one you can move.
First move the upper point vertically. The graph changes height, but its tangent slope at the selected input does not change. Next move the lower point across zero. Every member decreases where \(2x<0\) and increases where \(2x>0\), so each has a minimum at \(x=0\). Their minimum heights can differ.
Giving the family a name
For example,
\[\int 2x\,dx=x^2+C.\]The right side represents a family of functions. A particular choice, such as \(C=3\), selects one member. There are no endpoints in this notation; definite integrals and their connection with area come later.
Reading familiar derivative rules backwards
For a real exponent \(n\ne-1\), the power rule gives
\[\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n.\]Consequently,
\[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\ne-1.\]Use this on an interval where the powers are real and the derivative identity holds. For arbitrary real exponents, \(x>0\) is a safe domain; nonnegative integer powers also work on all of \(\mathbb R\). Negative powers require \(x\ne0\).
The sum and constant-multiple derivative rules also work backwards: if \(F^{\prime}=f\) and \(G^{\prime}=g\) on the same interval, then
\[\begin{aligned} &\int\bigl(af(x)+bg(x)\bigr)\,dx\\ &\qquad=aF(x)+bG(x)+C \end{aligned}\]for constants \(a,b\). Differentiating the right side verifies the statement. One arbitrary constant is enough: a sum of constant terms is still a constant.
Example 1
Recovering a polynomial
Find all antiderivatives of \(3x^2-4x+2\) on \(\mathbb R\).
Treat each term separately. To obtain \(3x^2\) after differentiating, use \(x^3\); to obtain \(-4x\), use \(-2x^2\); and to obtain the constant \(2\), use \(2x\). Thus
\[\begin{aligned} &\int(3x^2-4x+2)\,dx\\ &\qquad=x^3-2x^2+2x+C. \end{aligned}\]The check is immediate:
\[\begin{aligned} &\frac{d}{dx}(x^3-2x^2+2x+C)\\ &\qquad=3x^2-4x+2. \end{aligned}\]The final \(2x\) matters: a constant integrand has a linear antiderivative. Only an additive constant in the answer disappears during the check.
Example 2
A chain-rule factor must be accounted for
Find an antiderivative of \(e^{2x}+\cos(x)\) on \(\mathbb R\).
The derivative of \(e^{2x}\) is \(2e^{2x}\), twice the first term we want. Multiplying by \(1/2\) corrects that factor. Since the derivative of \(\sin(x)\) is \(\cos(x)\), we obtain
\[F(x)=\frac12e^{2x}+\sin(x).\]Indeed,
\[\begin{aligned} F^{\prime}(x)&=\frac12\cdot2e^{2x}+\cos(x)\\ &=e^{2x}+\cos(x). \end{aligned}\]This supplies one antiderivative. Adding \(C\) gives all of them. Recognizing a familiar function is a starting point; differentiating checks that every factor is correct.
The exceptional power and the role of the domain
At \(n=-1\), the reverse power formula would divide by zero. Instead, recall the logarithmic derivative:
\[\frac{d}{dx}\ln(x)=\frac1x\qquad(x>0).\]On \(x<0\), the real expression \(\ln(-x)\) has derivative
\[\frac{d}{dx}\ln(-x)=\frac1{-x}\cdot(-1)=\frac1x.\]These combine into the useful notation
\[\int\frac1x\,dx=\ln\lvert x\rvert+C,\]on any interval lying entirely on one side of zero. The absolute value makes the logarithm’s argument positive; it does not allow the interval to pass through zero.
One known value selects one member
A condition such as \(F(a)=b\) is an initial condition. It specifies a value of the unknown function; the input \(a\) need not be zero or represent a starting time.
If \(A\) is one antiderivative of \(f\) on an interval containing \(a\), all the possibilities are \(F(x)=A(x)+C\). Substituting the known value gives \(C=b-A(a)\). Therefore the unique member satisfying the condition is
\[F(x)=A(x)+b-A(a).\]This explains uniqueness when an antiderivative exists. It does not yet establish that every function has an antiderivative.
Example 3
A condition away from zero
Find \(F\) if \(F^{\prime}(x)=6x-4\) on \(\mathbb R\) and \(F(2)=7\).
First recover the family:
\[F(x)=3x^2-4x+C.\]Now impose the condition:
\[7=F(2)=12-8+C,\]so \(C=3\) and \(F(x)=3x^2-4x+3\). Check both pieces of information: its derivative is \(6x-4\), and its value at \(2\) is \(7\). The given value \(7\) is not itself the integration constant.
Example 4
From acceleration to position
A cart moves along a straight track with constant acceleration \(2\,\mathrm{m/s^2}\). At time \(t=0\) its velocity is \(3\,\mathrm{m/s}\) and its position is \(5\,\mathrm m\). Find its position at \(t=4\,\mathrm s\), assuming this model holds throughout \(0\le t\le4\).
Write \(t\) for the numerical time in seconds, \(v(t)\) for velocity in metres per second, and \(s(t)\) for position in metres. Since \(v^{\prime}(t)=2\), the velocity family is
\[v(t)=2t+C_1.\]The condition \(v(0)=3\) gives \(C_1=3\). Next, \(s^{\prime}(t)=v(t)=2t+3\), so
\[s(t)=t^2+3t+C_2.\]The condition \(s(0)=5\) gives \(C_2=5\). Hence
\[s(4)=16+12+5=33.\]The cart is at position \(33\,\mathrm m\). Its displacement from the initial position is \(33-5=28\,\mathrm m\). Differentiating the position twice returns the prescribed acceleration; checking \(s(0)\) and \(s^{\prime}(0)\) returns both initial values.
Recovering position from acceleration introduced two independent constants. Each records information lost at a separate differentiation step.
A final check before moving on
When you find an indefinite integral, differentiate your answer, check that it is defined on the intended interval, and include the arbitrary constant when the full family is requested. If a value is prescribed, check that condition as well.
We now know how a rate can determine a function up to a constant. In the next lesson, we approach accumulation by adding small pieces of area. The connection between these two ideas will become the fundamental theorem of calculus.