Finding Particular Polynomial Antiderivatives

From a Derivative to One Particular Function

A polynomial derivative describes how a function changes, but it does not identify the function completely. If two functions have the same derivative, they can differ by a constant. A specified function value supplies the information needed to select one function from that family.

This tutorial focuses on initial-value problems where the derivative is a polynomial. The process is direct: find the general antiderivative by integrating term by term, then use the given value of the function to determine the constant.

For this topic, \(p(x)\) is a polynomial and the function is considered on the real numbers. The value condition \(y(a)=b\) is also called an initial condition. The word “initial” does not require that \(a\) be zero or that it be the smallest input under consideration; it simply identifies the point where the function value is prescribed.

The General Calculation

Write the polynomial derivative in powers of \(x\), including any constant term. For example, if \(p(x)=c_nx^n+\cdots+c_1x+c_0\), where the \(c_k\) are constants, integrate each term. The power rule gives an antiderivative for each term, and linearity allows the terms to be combined.

Here \(C\) is an arbitrary constant. It appears because the derivative of a constant is zero: adding any constant to an antiderivative leaves its derivative unchanged. The condition \(y(a)=b\) determines which value of \(C\) gives the required function.

The formula includes \(k=0\): a constant derivative term \(c_0\) contributes \(c_0x\) to the antiderivative. It does not contribute another constant in place of \(c_0x\). Keeping that distinction clear prevents a frequent error.

A Reliable Step-by-Step Method

The final check has two parts because the problem imposes two conditions. Differentiating confirms the derivative condition; evaluating at \(a\) confirms the value condition. Neither check alone verifies both requirements.

Worked Examples

Why the Initial Condition Determines One Solution

The method works because the derivative determines an antiderivative up to an additive constant, and the value condition fixes that constant. The argument can be stated directly for any polynomial \(p\).

To establish existence, define \(P(x)=\sum_{k=0}^{n}\frac{c_k}{k+1}x^{k+1}\). The power rule gives \(P'(x)=p(x)\). Set \(C=b-P(a)\) and define \(y(x)=P(x)+C\). Then \(y'(x)=P'(x)=p(x)\), while \(y(a)=P(a)+b-P(a)=b\). Thus a solution exists.

For uniqueness, suppose \(y_1\) and \(y_2\) both satisfy the two conditions. Their difference has derivative \((y_1-y_2)'=p-p=0\), so it is constant. Since \(y_1(a)=b=y_2(a)\), that constant is \(y_1(a)-y_2(a)=0\). Therefore \(y_1-y_2=0\) everywhere, which means \(y_1=y_2\). Hence there is exactly one solution.

This reasoning also explains why one function value is exactly the information needed here. The family of antiderivatives differs only by one constant, and one condition fixes that one unknown.

A Common Mistake: Forgetting to Evaluate at the Given Input

A frequent error is to set the integration constant equal to the prescribed function value. That is only correct when the other terms of the antiderivative evaluate to zero at the specified input.

For example, return to \(y'(x)=6x^2-4x+3\), \(y(1)=5\). A wrong calculation might be

followed by the proposed answer \(y(x)=2x^3-2x^2+3x+5\). But substituting \(x=1\) shows the problem:

The correction is to include the value of every \(x\)-dependent term before solving for \(C\): \(5=2-2+3+C=3+C\), so \(C=2\). The prescribed value belongs to the whole function at \(x=a\), not to the constant alone.

Why This Method Is Useful

Finding a particular antiderivative is more informative than writing down an indefinite integral alone. The general antiderivative describes all functions with the specified polynomial derivative; an initial condition selects the one function that also passes through a prescribed point. This makes the method useful whenever a rate of change is known together with a function value.

For accuracy, preserve the order of operations when substituting a negative input, include the sign of every coefficient, and delay simplifying until each term has been evaluated. Finally, check the result against both conditions. A correct-looking integration step can still produce the wrong particular function if the constant is determined incorrectly.