The claim is F(x) = x^3. Its height at x = 1 is F(1) = 1, the same as f(1) = 1. Which statement is correct?

F is not an antiderivative, because F'(1) = 3 and f(1) = 1

F is an antiderivative, because F(1) = f(1)

F is an antiderivative, because x^3 is a power of x

Check an antiderivative

Builds on determining a particular antiderivative. 7 minutes.

Differentiate the claim

A student says that \(F(x) = x^{3/2}\) is an antiderivative of \(f(x) = \sqrt{x}\). The top graph shows \(f\) in chalk, and the lower graph shows the claim in teal.

To test the claim, differentiate it. The red curve on the top graph shows \(F'(x)\), the slope of the claim at each \(x\). The claim is correct only if the red curve lies on the chalk curve.

The red curve is above the chalk curve. At \(x = 4\) the chalk curve has height \(\cy{2}\), but \(F'(4) = \cr{3}\), so the claim is 1.5 times too steep. The short dashes along the teal curve have the slope \(f\) asks for, and they are red because the claim cuts across them.

Divide the claim by 1.5, with the slider or the button. The red curve moves down onto the chalk curve and disappears under it, and the dashes turn orange. The claim \(\cb{\tfrac23 x^{3/2}}\) has derivative \(\sqrt{x}\), so it passes.

Compare the slope with f, not the height

A claimed antiderivative F passes the test when its slope F′ equals f at every x. The red curve on top shows F′ wherever it leaves the chalk curve.

Press "Use a wrong claim". The red curve leaves the chalk curve, so the claim fails. Press "Use the correct claim" to return.

Press "Raise the claim". The teal curve moves up and no red curve appears, because a shift changes the heights of F and not its slopes.

The test cannot find the constant, so if a claim passes, the claim plus any constant passes too. It compares the slope of F with f, never the height of F.

Notation

To test a claimed antiderivative \(F\) of \(f\), differentiate \(F\).
Compare \(F'(x)\) with \(f(x)\). They must be equal for every \(x\) in the interval.
\[\cb{F}'(x) \;\overset{?}{=}\; \cy{f(x)}\]
Differentiate \(F\) with care. A missed chain rule or product rule factor is the usual error.
If \(F\) passes, then \(F + C\) passes for every \(C\), because the derivative of \(C\) is 0.

When it fails

Change \(f\) to 4 with the button. The claim stays \(F(x) = x^2\). At \(x = 2\), the slope \(F'(2) = 4\) equals \(f(2) = 4\), so a test at that one input passes.

Move the slider to \(x = 1\). There \(F'(1) = 2\), but \(f(1) = 4\). The red curve meets the chalk line at \(x = 2\) only. A test must hold for every \(x\), so compare the formulas \(2x\) and 4, not two values.

Example 1: A claim that passes

A student claims that \(F(x) = 2x^3 - 4x^2 + 9\) is an antiderivative of \(f(x) = 6x^2 - 8x\). Test the claim.

Differentiate each term of \(F\). The constant 9 has derivative 0.
\[F'(x) = 6x^2 - 8x + 0\]
Compare with \(f\). The two expressions are the same for every \(x\), so no red curve appears and the orange dashes lie along the teal curve.
The claim passes. The constant 9 raises the teal curve and does not change its slope. The claim now moves down by 9, and the dashes still lie along it.

Example 2: A claim that fails

A student claims that \(F(x) = e^{3x}\) is an antiderivative of \(f(x) = e^{3x}\). Test the claim.

Differentiate \(F\). The function \(e^{3x}\) has an inside function, \(3x\), so the chain rule applies.
A tempting wrong move: The derivative of \(e^{3x}\) is \(e^{3x}\), so the claim passes.
\[F'(x) = e^{3x}\]
The red curve is above the chalk curve. At \(x = 0\) the chalk curve has height 1, but the red curve has height 3, because the chain rule adds a factor.
Use the chain rule. The inside function \(3x\) has derivative 3, so the result is multiplied by 3.
\[F'(x) = 3e^{3x}\]
The result is 3 times \(f\), so the claim fails. Divide the claim by 3 and differentiate again. This time the result is \(f\), and the red curve disappears under the chalk curve.
\[F(x) = \tfrac13 e^{3x} \;\Rightarrow\; F'(x) = e^{3x}\]

Practice

A student claims that \(F(x) = \dfrac{(x^2 + 1)^3}{3}\) is an antiderivative of \(f(x) = 2x(x^2 + 1)^2\). Test the claim.

Differentiate with the chain rule, and compare with the picture. If the claim is correct, no red curve appears and the orange dashes lie along the teal curve.

Solution

The outer function is a cube, and the inside function \(x^2 + 1\) has derivative \(2x\).
\[F'(x) = \frac{1}{3}\cdot 3(x^2 + 1)^2 \cdot 2x\]
The 3s cancel.
\[F'(x) = 2x(x^2 + 1)^2\]
This equals f, so the claim passes.

Questions

Questions 1 and 2 use \(f(x) = x^2\) in chalk. The teal curve below is a claimed antiderivative \(F\). The red curve on the top graph shows \(F'(x)\) wherever it differs from \(f\).

The claim is \(F(x) = x^3\). Its height at \(x = 1\) is \(F(1) = 1\), the same as \(f(1) = 1\). Which statement is correct?
- \(F\) is not an antiderivative, because \(F'(1) = 3\) and \(f(1) = 1\)
- \(F\) is an antiderivative, because \(F(1) = f(1)\)
- \(F\) is an antiderivative, because \(x^3\) is a power of \(x\)
Answer
\(F\) is not an antiderivative, because \(F'(1) = 3\) and \(f(1) = 1\)
The claim is \(F(x) = x^2\). The red curve meets the chalk curve at one positive \(x\). Which \(x\)?
- 2
- 1
- 3
Answer
2
Which function \(F\) has \(F'(x) = x\sin x\)?
- \(\sin x - x\cos x\)
- \(-x\cos x\)
- \(\dfrac{x^2}{2}\sin x\)
Answer
\(\sin x - x\cos x\)

Summary

You differentiate the claimed antiderivative and compare the result with f at every x.

This works because an antiderivative is defined by its derivative, so a claim passes only when its slope equals the height of f everywhere.

You use it to test any antiderivative you find, and the test cannot detect the constant C.

This is one of five skills in topic 6.8 of Which & Why: AP Calculus.