Tutorials › Calculus I › Antiderivatives And Reverse Differentiation

Antiderivatives and Basic Integrals · Tutorial 1046 of 1495

Antiderivatives And Reverse Differentiation

An antiderivative reverses the question asked by differentiation: check a proposed function by finding its derivative at every point of the interval.

Beginner 8 min read

What You'll Learn

  • Define an antiderivative of a function on an interval.
  • Identify the interval on which a proposed antiderivative must work.
  • Verify a candidate by differentiating and comparing functions pointwise.
  • Use the derivative definition to explain why this check is decisive.
  • Recognize common errors, including checking only selected input values.

From Differentiation to Its Reverse

Differentiation describes how a function changes. If a function \(F\) has derivative \(f\), then \(f\) gives the rate of change of \(F\) at each input. This connection also suggests a question in the opposite direction: if a function \(f\) is given, can we find a function whose derivative is \(f\)?

For example, the derivative of \(x^2\) is \(2x\). This means that \(x^2\) is one function whose derivative is \(2x\). The same idea applies to less immediate cases: propose a function, differentiate it, and check whether the result is the function required.

This reverse viewpoint is useful because it turns a proposed answer into something that can be tested directly. The test is not whether the proposed function looks plausible or has convenient values. The test is whether its derivative agrees with the target function throughout the interval under consideration.

Definition of an Antiderivative

Definition: Let \(I\) be an open interval, and let \(f:I\to\mathbb{R}\) be a function. A function \(F:I\to\mathbb{R}\) is an antiderivative of \(f\) on \(I\) if \(F\) is differentiable at every \(x\in I\) and $$ F'(x)=f(x) $$ for every \(x\in I\).

The definition has two requirements. First, \(F\) must be differentiable throughout the interval. Second, its derivative must equal \(f\) at every input in that interval. A match at just one point, or at a selection of points, is not enough.

The interval is part of the claim. Saying that \(F\) is an antiderivative of \(f\) on \(I\) means the derivative equality has been verified for all inputs in \(I\). A candidate might work on one interval but fail to be defined or differentiable on a larger one. For that reason, always keep the domain and the interval in view while checking.

The definition asks for a relationship between two functions: the derivative \(F'\) and the target function \(f\). It does not say that the functions \(F\) and \(f\) themselves must be equal. In fact, the required comparison is between \(F'\) and \(f\).

How the Check Works

To check a proposed antiderivative, calculate the derivative of the proposed function using the differentiation rules that apply, then compare the result with the target function. If the expressions agree for every input in the interval, the definition is satisfied. If they disagree at even one input, the candidate is not an antiderivative of that target function on the interval.

1
State the target and interval.
Write down which function \(f\) the candidate must have as its derivative and where the claim is supposed to hold.
2
Differentiate the candidate.
Find \(F'(x)\), showing the steps needed to justify the calculation.
3
Compare the functions.
Check whether \(F'(x)=f(x)\) for every \(x\) in the stated interval, and then state the conclusion.

The reason this procedure is decisive is built into the definition of derivative. At an input \(x\), the derivative of \(F\), when it exists, is the limit of its difference quotient:

$$ F'(x)=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}. $$

Consequently, if the derivative calculation gives \(F'(x)=f(x)\) at every \(x\in I\), then at every such input the rate of change of \(F\), as measured by this limit, is exactly \(f(x)\). This is precisely the condition in the definition. Conversely, if \(F\) is an antiderivative of \(f\) on \(I\), the definition itself requires this equality at every point of \(I\). Thus differentiation is not merely a helpful test: it checks the defining condition.

Worked Checks

Worked Example: A Polynomial Candidate

Check whether \(F(x)=3x^2-4x+7\) is an antiderivative of \(f(x)=6x-4\) on \(\mathbb{R}\).

The proposed function is a polynomial, so it is differentiable for every real \(x\). Differentiate each term:

$$ F'(x)=\frac{d}{dx}(3x^2)-\frac{d}{dx}(4x)+\frac{d}{dx}(7) =3(2x)-4(1)+0 =6x-4. $$

The required function is \(f(x)=6x-4\), so \(F'(x)=f(x)\) for every \(x\in\mathbb{R}\). The candidate is therefore an antiderivative of \(f\) on \(\mathbb{R}\).

As a direct check at \(x=2\), \(F'(2)=6(2)-4=8\), and \(f(2)=6(2)-4=8\). This value agrees, while the full derivative calculation establishes the equality for all real inputs.

Worked Example: A Trigonometric Candidate

Check whether \(F(x)=\sin x\) is an antiderivative of \(f(x)=\cos x\) on \(\mathbb{R}\).

The sine function is differentiable for all real \(x\). Using its derivative,

$$ F'(x)=\frac{d}{dx}(\sin x)=\cos x. $$

Since the target is \(f(x)=\cos x\), this gives \(F'(x)=f(x)\) for every real \(x\). Thus \(\sin x\) is an antiderivative of \(\cos x\) on \(\mathbb{R}\).

For a numerical check at \(x=0\), \(F'(0)=\cos 0=1\), and \(f(0)=\cos 0=1\). Again, this single value illustrates the equality but does not replace the derivative identity, which holds throughout the interval.

Worked Example: Applying the Chain Rule

Check whether \(F(x)=e^{2x}\) is an antiderivative of \(f(x)=2e^{2x}\) on \(\mathbb{R}\).

The exponential function is differentiable for every real input. Since the exponent \(2x\) has derivative \(2\), the chain rule gives

$$ F'(x)=\frac{d}{dx}\left(e^{2x}\right) =e^{2x}\cdot\frac{d}{dx}(2x) =e^{2x}\cdot 2 =2e^{2x}. $$

The target function is \(f(x)=2e^{2x}\), exactly the derivative found. Therefore \(F\) is an antiderivative of \(f\) at every real input.

At \(x=0\), the derivative is \(F'(0)=2e^0=2\), and the target is \(f(0)=2e^0=2\), providing a further pointwise check.

A Derivation of the Verification Criterion

The test can be written as an exact equivalence. Suppose \(F:I\to\mathbb{R}\) is differentiable throughout an open interval \(I\), and \(f:I\to\mathbb{R}\). Then \(F\) is an antiderivative of \(f\) on \(I\) if and only if its derivative agrees with \(f\) at every point of \(I\).

1
Assume the derivative agrees everywhere.
Suppose \(F'(x)=f(x)\) for each \(x\in I\). The function \(F\) is differentiable throughout \(I\) by assumption, so both requirements in the definition of antiderivative hold. Hence \(F\) is an antiderivative of \(f\) on \(I\).
2
Assume \(F\) is an antiderivative.
By definition, \(F\) is differentiable at every \(x\in I\), and \(F'(x)=f(x)\) for every \(x\in I\). Thus the derivative equality must hold everywhere on the interval.

The two directions show why the check is both sufficient and necessary. A correct derivative calculation proves the proposed relationship; a mismatch means the defining relationship fails. There is no additional visual or numerical test required.

A Common Mistake: Matching Only Part of the Target

A candidate can produce much of the desired derivative and still fail. Suppose the target is \(f(x)=3x^2+1\), and someone proposes \(F(x)=x^3\). The calculation is

$$ F'(x)=\frac{d}{dx}(x^3)=3x^2. $$

This derivative is missing the \(+1\), so it is not equal to \(f(x)=3x^2+1\). For example, at \(x=0\), \(F'(0)=0\), whereas \(f(0)=1\). The proposed function is not an antiderivative of the target on \(\mathbb{R}\).

One correction is to propose \(G(x)=x^3+x\). Checking it gives

$$ G'(x)=\frac{d}{dx}(x^3)+\frac{d}{dx}(x) =3x^2+1 =f(x). $$

The corrected candidate passes the check at every real input. The comparison must include the entire target function, not just its leading term or the part that is easiest to recognize.

Why the Interval and Every Input Matter

An equality of functions on an interval means equality at each input in that interval. Checking a handful of input values can catch an error, but agreement at those values alone does not establish the required identity. A derivative formula that matches \(f\) only at \(x=0\), for example, does not meet the definition unless it matches at every point in \(I\).

It is also important not to lose track of where the candidate is defined and differentiable. The definition applies throughout the stated interval. When using standard differentiation rules, verify that the candidate is defined and differentiable there before concluding that it is an antiderivative.

Thinking in terms of rates of change helps interpret the definition: \(f(x)\) specifies the rate that the proposed function \(F\) must have at input \(x\). Differentiating \(F\) tests that rate directly. The earlier material on accumulation also gives a useful context for rates and accumulated change, but the test here is simply the derivative equality required by the definition.

Summary

An antiderivative of \(f\) on an interval \(I\) is a function \(F\) that is differentiable throughout \(I\) and satisfies \(F'(x)=f(x)\) for every \(x\in I\). To check a proposed candidate, differentiate it, compare the result with the target function, and confirm that the equality holds throughout the stated interval.

Key takeaway: A proposed antiderivative is verified by differentiation. The check succeeds exactly when the derivative equals the required function at every point of the interval.

Check Your Understanding

For each question, pay attention to the candidate, the target function, and the interval.

  1. State the definition of an antiderivative of \(f\) on an open interval \(I\).
  2. Is \(F(x)=x^2+3x\) an antiderivative of \(f(x)=2x+3\) on \(\mathbb{R}\)? Show the derivative check.
  3. Is \(F(x)=\cos x\) an antiderivative of \(f(x)=-\sin x\) on \(\mathbb{R}\)? Explain using differentiation.
  4. Why is agreement between \(F'(x)\) and \(f(x)\) at a few selected inputs not enough to establish that \(F\) is an antiderivative on an interval?
  5. A proposed candidate has derivative \(4x^3\), while the target is \(4x^3-2\). Does it pass the check? State why.

Test yourself: Calculus I final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon