From an Antiderivative Family to One Function
An indefinite integral describes a family of functions. If \(F_0\) is an antiderivative of \(f\) on an interval, then every function \(F_0(x)+C\), where \(C\) is a constant, has the same derivative \(f\). The constant shifts the graph vertically without changing its derivative.
A specified function value selects one member of this family. For example, if we know both that \(F'(x)=f(x)\) and that \(F(a)=b\), then the second condition tells us which vertical shift to use. This is an initial condition: it gives the value of the unknown function at a specified input.
The derivative condition determines the antiderivative family; the initial condition determines the constant in that family. We will make this relationship precise and prove that, when an antiderivative exists on an interval, exactly one member of the family satisfies any specified value at a point in that interval.
How the Initial Condition Determines the Constant
Suppose we already know one antiderivative \(F_0\) of \(f\) on an interval \(I\). The general antiderivative is \(F_0(x)+C\). To impose \(F(a)=b\), substitute \(x=a\) and set the result equal to \(b\):
Solving this equation gives \(C=b-F_0(a)\). Thus the particular antiderivative is
This formula is useful even when \(F_0(a)\) is not zero: evaluate the known antiderivative at the specified input, then choose the constant that makes the output equal the specified value. The point \(a\) must lie in the domain where the antiderivative is defined.
Worked Example: A Logarithmic Antiderivative
Find \(F\) satisfying \(F'(x)=1/x\) for \(x>0\) and \(F(2)=5\).
On the interval \((0,\infty)\), an antiderivative of \(1/x\) is \(\ln x+C\). Apply the initial condition by substituting \(x=2\):
Therefore, \(C=5-\ln 2\), and the particular antiderivative is
Check the derivative: \(F'(x)=1/x\) for \(x>0\). Check the initial condition: \(F(2)=\ln 2+5-\ln 2=5\). Both conditions hold.
Existence and Uniqueness on an Interval
The method is more than a convenient calculation. It gives a general result. Assume \(f\) has at least one antiderivative \(F_0\) on an interval \(I\). The earlier result that antiderivatives on an interval differ by a constant means every antiderivative has the form \(F_0+C\). We can now establish both existence and uniqueness for the specified value.
Here, “exactly one” has two parts. Existence means that at least one such function can be constructed. Uniqueness means that no two different antiderivatives on \(I\) can both meet the same initial condition.
Let \(F_0\) be any antiderivative of \(f\) on \(I\), so \(F_0'(x)=f(x)\).
Set \(C=b-F_0(a)\), and define \(F(x)=F_0(x)+C\). Then \(F'(x)=F_0'(x)=f(x)\), and \(F(a)=F_0(a)+b-F_0(a)=b\). This proves existence.
If \(G\) is another solution, then \(G-F\) has derivative \(G'(x)-F'(x)=f(x)-f(x)=0\) on \(I\). A function with derivative zero throughout an interval is constant there, so \(G-F\) is constant.
At \(x=a\), \(G(a)-F(a)=b-b=0\). The constant \(G-F\) must therefore be zero everywhere on \(I\), so \(G=F\). This proves uniqueness.
The interval matters in the uniqueness argument: a function whose derivative is zero is constant on an interval. If the domain is disconnected, it may have different constants on separate pieces, and a value on one piece cannot determine its value on another.
Working with Other Antiderivative Families
The same procedure applies whether the family comes from a constant function, a power, or another function whose antiderivative is known. The steps remain the same: write the general family, substitute the specified input, solve for the constant, and verify both conditions.
Worked Example: A Fractional-Power Antiderivative
Find \(G\) satisfying \(G'(x)=1/\sqrt{x}\) for \(x>0\) and \(G(9)=1\).
Since \(1/\sqrt{x}=x^{-1/2}\), the power rule for antiderivatives gives the family \(G(x)=2\sqrt{x}+C\) on \((0,\infty)\). Apply the condition at \(x=9\):
It follows that \(C=-5\), so
Check the derivative: \(\frac{d}{dx}(2\sqrt{x}-5)=1/\sqrt{x}\) for \(x>0\). Check the value: \(G(9)=2(3)-5=1\). The function satisfies both conditions.
Worked Example: A Constant Derivative
Find \(H\) satisfying \(H'(x)=3\) for all real \(x\) and \(H(-1)=4\).
An antiderivative of the constant function \(3\) is \(3x+C\), so write \(H(x)=3x+C\). Substitute the specified input and value:
Adding \(3\) to both sides gives \(C=7\). Hence \(H(x)=3x+7\).
Check the derivative: \(H'(x)=3\). Check the initial condition: \(H(-1)=3(-1)+7=-3+7=4\). Both requirements are satisfied.
Why One Value Is Enough
Every member of an antiderivative family has the same derivative because adding a constant does not affect differentiation. But different constants give different function values at every point: if \(F_1(x)=F_0(x)+C_1\) and \(F_2(x)=F_0(x)+C_2\), then \(F_1(a)-F_2(a)=C_1-C_2\). Therefore, if both functions have the same value at \(a\), then \(C_1-C_2=0\), so the constants are equal.
This explains why one initial value is exactly what is needed to fix one additive constant. It does not provide a new derivative rule; it selects one function from a family already obtained by antiderivatives. If no antiderivative exists on the domain under consideration, this construction cannot be used there. The theorem assumes an antiderivative exists.
Common Mistake: Confusing the Input and the Function Value
In \(F(a)=b\), \(a\) is the input at which the function is evaluated, and \(b\) is the output. A common error is to treat the specified value \(b\) as the input when substituting into the antiderivative.
For example, for \(F'(x)=1/x\) and \(F(2)=5\), the incorrect substitution \(F(5)=\ln 5+C=5\) uses \(b=5\) as the input. That solves a different condition, namely \(F(5)=5\). The correct substitution uses \(a=2\):
Always identify the input and output from the condition before substituting. After determining the constant, check the original value condition, not a newly created one.
When the Domain Is Disconnected
The interval assumption in the uniqueness theorem cannot simply be discarded. For example, let the domain be \(I=(-\infty,0)\cup(0,\infty)\) and consider \(F'(x)=0\) there with \(F(1)=2\). A function satisfying these conditions can equal \(2\) for \(x>0\) and any constant for \(x<0\). Its derivative is zero on both pieces, and the condition at \(1\) gives no information about its values on the negative piece.
Thus a single initial condition fixes the constant on the interval containing its input. When the domain has separate intervals, each component may have its own constant; additional conditions may be needed to determine those other constants.
Check Your Understanding
For each question, identify the derivative condition and the specified value before solving for a constant.
- For \(F'(x)=1/x\) on \(x>0\) and \(F(1)=3\), what is \(F(x)\)?
- For \(G'(x)=1/\sqrt{x}\) on \(x>0\) and \(G(4)=0\), what constant is required in \(G(x)=2\sqrt{x}+C\)?
- Why does substituting the specified input into the general antiderivative determine the constant?
- What two checks should be made after finding a particular antiderivative?
- Why might one initial condition fail to determine an antiderivative on a disconnected domain?
