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Undoing differentiation, why the constant is a theorem, the one exponent the power rule misses, and what has no antiderivative at all.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find antiderivatives by reversing the differentiation rules and check each by differentiating, explain why two antiderivatives of the same function differ by a constant and which theorem that rests on, say why the hypothesis of a single interval matters, handle the $n = -1$ exception with the absolute value the domain requires, use an initial condition to pick one antiderivative out of the family, and say why some elementary functions have no elementary antiderivative.
$F$ is an antiderivative of $f$ on an interval if $F' = f$ there.
Antiderivatives are never unique, because differentiating destroys constants: if $F' = f$ then $(F + C)' = f$ for every constant $C$. What is worth knowing is that nothing else works.
Theorem. If $F' = G' = f$ on an interval, then $F - G$ is constant.
Proof. $(F-G)' = 0$ throughout, and a function with zero derivative on an interval is constant — by the mean value theorem. $\square$
So the antiderivatives of $f$ are exactly $\{F + C\}$, and the $+C$ is a theorem, not a convention. The word interval is a hypothesis: on a disconnected domain the pieces can carry different constants.
The indefinite integral $\int f(x)\,dx$ denotes that whole family.
| $f$ | $\int f\,dx$ |
|---|---|
| $x^n$, $n \ne -1$ | $\dfrac{x^{n+1}}{n+1} + C$ |
| $x^{-1}$ | $\ln\|x\| + C$ |
| $e^x$ | $e^x + C$ |
| $\cos x$ | $\sin x + C$ |
| $\sin x$ | $-\cos x + C$ |
| $\sec^2 x$ | $\tan x + C$ |
Every row is a derivative from unit 3, read right to left. There is nothing new to learn here — only a table to read in the other direction, and the discipline of checking by differentiating.
Another way: picture
A family of curves stacked vertically, each a copy of the others shifted up or down. At any $x$ they all have the same slope, so they all have the same derivative. An initial condition is a single point in the plane, and exactly one curve of the family passes through it.
Another way: steps
The reverse power rule divides by $n+1$, so it fails at exactly one exponent: $n = -1$. And that gap is not an inconvenience to be worked around — it is where the natural logarithm enters the subject.
$$\int \frac{1}{x}\,dx = \ln|x| + C.$$
The absolute value is not decoration. $1/x$ is defined for negative $x$ and $\ln x$ is not; $\ln|x|$ has derivative $1/x$ on both halves of the domain.
And the domain really is in two halves. Strictly, $$\int \frac{1}{x}\,dx = \begin{cases}\ln x + C_1 & x > 0\\ \ln(-x) + C_2 & x < 0\end{cases}$$ with two independent constants, because the theorem above needs an interval and $x \ne 0$ is not one. Almost every textbook writes a single $C$, and almost every problem stays on one side of the origin, so the shortcut is harmless in practice and wrong in principle — which is worth knowing before meeting an improper integral across a pole.
Forgetting $+C$. It costs marks and, more seriously, it loses the family. An initial value problem has nothing to solve without it.
Antidifferentiating a product by antidifferentiating the factors. There is no product rule in reverse that is this simple. $\int x\cos x\,dx \ne \frac{x^2}{2}\sin x$. Differentiating the right side proves it immediately, which is why the check matters.
Using the power rule at $n = -1$. It divides by zero.
Antidifferentiating a quotient as a quotient. $\int \dfrac{1}{x^2+1}dx$ is $\arctan x + C$ and nothing about the shape of the integrand suggests it. Antidifferentiation is recognition, not procedure.
Expecting every elementary function to have an elementary antiderivative. $e^{-x^2}$, $\dfrac{\sin x}{x}$ and $\sqrt{1+x^4}$ have none — proved impossible, not merely unknown. Differentiation always terminates; antidifferentiation often does not, and that asymmetry is the deepest fact in this lesson.
Differentiation is an algorithm. Any elementary function can be differentiated by applying the rules mechanically, and the answer is always elementary. There is no judgement in it.
Antidifferentiation is recognition. There is no procedure that works in general, the rules run backwards only in special cases, and — the part that surprises — some elementary functions have no elementary antiderivative at all. $\int e^{-x^2}dx$ is not difficult; it is impossible, and that impossibility is a theorem (Liouville's), not a gap in anyone's technique.
This is why the integral needs its own definition rather than being defined as "whatever undoes the derivative". $\int_0^1 e^{-x^2}dx$ is a perfectly good number, and the next three lessons define it without ever finding an antiderivative — which is exactly what makes the fundamental theorem worth proving when the two ideas finally meet.
$\displaystyle\int \frac{3x^2 - \sqrt{x}}{x}\,dx$. Divide term by term first: $3x - x^{-1/2}$.
The power rule needs powers.
Antidifferentiate each: $\dfrac{3x^2}{2} - \dfrac{x^{1/2}}{1/2} = \dfrac{3x^2}{2} - 2\sqrt{x}$.
Raise, divide, term by term.
So the answer is $\dfrac{3x^2}{2} - 2\sqrt x + C$. Differentiate to check: $3x - x^{-1/2}$. Correct.
The check is one line.
$f'(x) = 6x^2 - 4$ with $f(1) = 5$. General antiderivative: $f(x) = 2x^3 - 4x + C$.
The family first.
Apply the condition: $2 - 4 + C = 5$, so $C = 7$.
One condition, one constant.
$f(x) = 2x^3 - 4x + 7$. One curve of the family, picked out by one point.
Term by term. $4x^3$ raises to $x^4$ and divides by $4$: $x^4$.
Reverse power rule.
$\sec^2 x$ is the derivative of $\tan x$, read off the table backwards.
$x^4 + \tan x + C$. Differentiate to check: $4x^3 + \sec^2 x$.
Match each integrand to the antiderivative rule it calls for.
| Use $x^{n+1}/(n+1)+C$ | Use $\ln|x|+C$ | Include an arbitrary constant | Subtract lower antiderivative value from upper value | |
|---|---|---|---|---|
| $x^n$, $n\ne-1$ | ||||
| $1/x$ | ||||
| An indefinite integral | ||||
| A definite integral evaluated by an antiderivative |
Match each integration fact to the conclusion it supports.
| The general answer contains an arbitrary constant | The condition fixes the constant in $F(x)=x^2+C$ | It equals $x^5/5+C$ | It equals $\ln|x|+C$ | |
|---|---|---|---|---|
| Both $x^2/2+C$ and $x^2/2+K$ differentiate to $x$ | ||||
| $F'(x)=2x$ and $F(3)=12$ | ||||
| $\int x^4\,dx$ | ||||
| $\int x^{-1}\,dx$ |
Find the general antiderivative of $35x^{4}$. Write the constant as $C$.
Answer:
Why do two antiderivatives of the same function differ by a **constant**, rather than by some other function?
| $F-G$ has derivative zero | Equals $F'-G'$ | Forces the function to be constant | Selects one member of the antiderivative family | |
|---|---|---|---|---|
| $F'=G'$ on one interval | ||||
| $(F-G)'$ | ||||
| A zero derivative throughout an interval | ||||
| One value condition $F(a)=b$ |
$F'(x) = 6x$ and $F(5) = 1$. What is the constant in $F(x) = 3x^2 + C$?
Answer:
A particle has acceleration $6$ m/s², starts at the origin with velocity $8$ m/s. Where is it after $4$ seconds?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The reverse power rule gives $\displaystyle\int x^n dx = \frac{x^{n+1}}{n+1} + C$. Which $n$ does it fail for, and what is the antiderivative there?
| Use $x^{n+1}/(n+1)+C$ | Use $\ln|x|+C$ | Vanishes when $n=-1$ | Requires absolute value in the logarithm | |
|---|---|---|---|---|
| $x^n$ with $n\ne-1$ | ||||
| $x^{-1}$ | ||||
| The factor $n+1$ in the reverse power rule | ||||
| Negative as well as positive $x$ |
You can antidifferentiate, check by differentiating, and fix the constant from a condition. Say in your own words why the $+C$ is a theorem rather than a convention. Next: Riemann sums, and an integral defined without any antiderivative at all.
8. Your turn: $\displaystyle\int (4x^3 + \sec^2 x)\,dx$, step 3