Single-Variable-Calculus · Unit 12 · Video 2 · Interactive Practice

Antiderivatives Differ Only by a Constant

IKey Formulas

Formula	Name	Meaning
$\displaystyle\int g(x)\,dx = G(x) + c$	Indefinite integral (antiderivative)	$G'(x) = g(x)$ , or $dG = g(x)\,dx$
$\displaystyle\int x^a\,dx = \frac{x^{a+1}}{a+1} + c$ , $a \neq -1$	Power rule read backwards	From $d(x^{a+1}) = (a+1)x^a\,dx$ , divided by $a+1$
$\displaystyle\int \frac{dx}{x} = \ln\|x\| + c$	The exception $a = -1$	Holds for $x > 0$ and for $x < 0$ , where $\ln\|x\| = \ln(-x)$
$F' = G'$ on an interval $\Rightarrow F = G + c$	Uniqueness up to a constant	$(F - G)' = 0$ , so $F - G$ is constant by the Mean Value Theorem

Key Insight: Adding a constant never changes a derivative, since $(c)' = 0$ . The Mean Value Theorem shows this is the only ambiguity: on an interval, the rate of change determines the function up to its starting value.

IIVisualization 1 — The Family $-\cos x + c$

Every curve $y = -\cos x + c$ has slope $\sin x$ at each $x$ ; only the vertical shift $c$ differs.

x₀ 1.00 c 1

💡 Challenge: find the one antiderivative of $\sin x$ whose graph passes through $(0, 1)$ . Its starting value fixes $c$ .

IIIVisualization 2 — Reading the Table Backwards

For each pair, the slope of $G$ at $x$ equals the height of $g$ at $x$ : that is what $\int g\,dx = G + c$ says.

g(x) → G(x) x 1.00

IVVisualization 3 — One Constant per Interval

Both branches of $\ln|x| + c$ have derivative $1/x$ , but $x = 0$ splits the domain into two intervals.

c₁ (x < 0) 0.5 c₂ (x > 0) −1

The theorem's proof compares points $a < b$ inside one interval: $f(b) = f(a) + f'(c)(b - a)$ with $f = F - G$ and $f' \equiv 0$ . No such comparison crosses $x = 0$ , so $c_1$ and $c_2$ are independent.

VQuiz Questions

Problem 1 · A Negative Power

Find $\displaystyle\int x^{-3}\,dx$ .

✅ Correct! With

a = -3

\dfrac{x^{-2}}{-2} + c

, and

\big(-\tfrac{1}{2}x^{-2}\big)' = x^{-3}

❌ That is the derivative.

-3x^{-4}

(x^{-3})'

; integrating goes the other way, raising the power by one.

❌ Wrong direction for the power. The antiderivative of

x^a

has power

a + 1 = -2

, not

a - 1 = -4

❌ Check the sign. The denominator is

a + 1 = -2

, so the coefficient is

-\tfrac{1}{2}

; differentiate

\tfrac{1}{2}x^{-2}

and you get

-x^{-3}

❌ Not quite. Use

\int x^a\,dx = \dfrac{x^{a+1}}{a+1} + c

with

a = -3

Show solution

Here $a = -3 \neq -1$ , so the power rule applies:

\int x^{-3}\,dx = \frac{x^{-3+1}}{-3+1} + c = \frac{x^{-2}}{-2} + c = -\frac{1}{2x^{2}} + c.

Check: $\dfrac{d}{dx}\Big(-\tfrac{1}{2}x^{-2}\Big) = -\tfrac{1}{2}\cdot(-2)x^{-3} = x^{-3}$ .

Problem 2 · The Negative Side of 1/x

Which function is an antiderivative of $\dfrac{1}{x}$ on the interval $x < 0$ ?

✅ Correct! For

x < 0

\ln|x| = \ln(-x)

, and the two minus signs in

\dfrac{1}{-x}\cdot(-1)

cancel to give

\dfrac{1}{x}

❌ Not defined there.

\ln x

only exists for

x > 0

; on

x < 0

the formula needs the absolute value,

\ln|x| = \ln(-x)

❌ One minus sign too many.

\big(-\ln(-x)\big)' = -\dfrac{1}{-x}\cdot(-1) = -\dfrac{1}{x}

. The chain rule already supplies the second minus sign.

❌ That is the derivative of $1/x$ .

\big(x^{-1}\big)' = -x^{-2}

; the question asks for a function whose derivative is

1/x

❌ Not quite. Differentiate each option on

x < 0

and look for

1/x

Show solution

The power rule fails at $a = -1$ (it would divide by $a + 1 = 0$ ), so use $\int \frac{dx}{x} = \ln|x| + c$ . For $x < 0$ , $|x| = -x$ . With $u = -x$ , $\frac{du}{dx} = -1$ :

\frac{d}{dx}\ln(-x) = \frac{1}{-x}\cdot(-1) = \frac{1}{x}.

So $\ln(-x)$ works. By contrast $\big(-\ln(-x)\big)' = -\tfrac{1}{x}$ , and $\ln x$ is undefined for $x < 0$ .

Problem 3 · Pinning Down the Constant

Given: $F'(x) = \sin x + x^2$ for all $x$ and $F(0) = 2$ . Find $F(x)$ .

✅ Correct!

F(0) = -\cos 0 + 0 + 3 = -1 + 3 = 2

, and

F'(x) = \sin x + x^2

❌ Check the starting value.

-\cos 0 = -1

, so this gives

F(0) = 1

, not

2

. The constant has to make up the

-1

❌ Sign of the cosine.

F(0) = 2

holds, but

(\cos x)' = -\sin x

. The antiderivative of

\sin x

-\cos x

❌ $2x$ is the derivative of $x^2$ .

F(0) = 2

holds, but

F'(x) = \sin x + 2

. The antiderivative of

x^2

\tfrac{1}{3}x^3

❌ Not quite. Antidifferentiate term by term, then choose

c

so that

F(0) = 2

Show solution

Step 1: every antiderivative. Reading the table backwards, term by term:

F(x) = -\cos x + \frac{x^3}{3} + c.

By the uniqueness theorem these are all the antiderivatives on $(-\infty, \infty)$ , so $F$ is one of them.

Step 2: the starting value.

F(0) = -\cos 0 + 0 + c = -1 + c = 2 \quad\Longrightarrow\quad c = 3.

F(x) = -\cos x + \tfrac{1}{3}x^3 + 3.

Check: $F'(x) = \sin x + x^2$ and $F(0) = -1 + 3 = 2$ .

Problem 4 · Same Derivative, Two Intervals

Given: $F(x) = \tan^{-1}x$ and $G(x) = -\tan^{-1}\!\big(\tfrac{1}{x}\big)$ , defined for $x \neq 0$ . Both have derivative $\dfrac{1}{1+x^2}$ wherever they are defined. What is $F(x) - G(x)$ ?

✅ Correct!

F - G

is constant on each interval, and the two constants differ:

\tfrac{\pi}{4} + \tfrac{\pi}{4} = \tfrac{\pi}{2}

x = 1

-\tfrac{\pi}{4} - \tfrac{\pi}{4} = -\tfrac{\pi}{2}

x = -1

❌ Equal derivatives do not make equal functions. They make

F - G

constant on an interval. At

x = 1

F - G = \tan^{-1}1 + \tan^{-1}1 = \tfrac{\pi}{2} \neq 0

❌ Right for $x > 0$ only. The domain

x \neq 0

is two intervals, so the constant may change across

0

. Evaluate at

x = -1

❌ The theorem applies on each interval.

(F - G)' = 0

x > 0

and on

x < 0

, so

F - G

is constant on each one.

❌ Not quite. Evaluate

F - G

at one point in each interval.

Show solution

Step 1: equal derivatives. By the chain rule,

G'(x) = -\frac{1}{1 + 1/x^2}\cdot\Big(-\frac{1}{x^2}\Big) = \frac{1}{x^2 + 1} = F'(x).

Step 2: apply the theorem on each interval. $(F - G)' = 0$ on $(0, \infty)$ and on $(-\infty, 0)$ , so $F - G$ is a constant on each, possibly a different one.

Step 3: one point per interval.

x = 1:\quad \tan^{-1}1 + \tan^{-1}1 = \tfrac{\pi}{4} + \tfrac{\pi}{4} = \tfrac{\pi}{2}

x = -1:\quad \tan^{-1}(-1) + \tan^{-1}(-1) = -\tfrac{\pi}{4} - \tfrac{\pi}{4} = -\tfrac{\pi}{2}

So $F - G = \tfrac{\pi}{2}$ for $x > 0$ and $-\tfrac{\pi}{2}$ for $x < 0$ . This is the same situation as $\ln|x| + c$ : one constant per interval.

Solved: 0 / 4

Antiderivatives Differ Only by a Constant

IKey Formulas

IIVisualization 1 — The Family −cos⁡x+c-\cos x + c−cosx+c

IIIVisualization 2 — Reading the Table Backwards

IVVisualization 3 — One Constant per Interval

VQuiz Questions

IIVisualization 1 — The Family $-\cos x + c$