4.10 Antiderivatives

Big idea

Every rule so far has run in one direction: given a function, produce its derivative. This section runs the process backward. Given a rate, recover a function that has that rate. The question is natural because rates are what the physical world hands you: a velocity, a growth rate, a marginal cost. What you want is the position, the population, the total cost.

Running a rule backward is harder than running it forward, and the asymmetry is worth noticing early. Differentiation is an algorithm; you can always do it. Antidifferentiation is a search; you recognize the answer or you do not. Every derivative rule you know becomes an antiderivative rule when read from right to left, which is why the list below is short and why the practice is mostly pattern recognition.

The second feature of the backward direction is that the answer is never unique. Constants differentiate to zero, so adding any constant to an antiderivative gives another one. The mean value theorem says that is the only freedom: two functions with the same derivative on an interval differ by a constant. So the full set of antiderivatives is one of them plus an arbitrary constant, and that arbitrary constant is exactly the information a single extra data point supplies.

That last observation makes these problems useful rather than merely formal. Knowing acceleration gives velocity up to a constant, and the initial velocity fixes the constant. Knowing velocity gives position up to a constant, and the initial position fixes that one. Two integrations and two initial conditions turn a law of motion into a trajectory.

Decoder

A function $F$ is an antiderivative of $f$ on an interval if $F'(x) = f(x)$ throughout it, and every antiderivative of $f$ on that interval has the form $F(x) + C$.

The phrase “on an interval” is doing real work. The claim that two antiderivatives differ by a constant comes from the mean value theorem, which needs a connected domain. On a domain split into pieces, such as the domain of $1/x$, each piece gets its own constant, and a first course usually suppresses that subtlety by working on one piece at a time.

The rest is a definition plus a uniqueness statement. The definition tells you how to check an answer: differentiate it and see whether you get back what you started with. That check is available on every problem in this section, it is quick, and it is the reason there is no excuse for a wrong antiderivative.

The classic mistake is dropping the constant. It costs nothing on a pure antidifferentiation exercise and costs everything on an initial value problem, where the constant is the whole point of the question being asked. The second classic mistake is inventing rules that do not exist. There is no product rule and no quotient rule for antiderivatives; the antiderivative of a product is not the product of the antiderivatives.

Definitions and results

Antiderivative. $F$ is an antiderivative of $f$ on an interval if $F'(x) = f(x)$ for every $x$ in that interval.

The general antiderivative. If $F$ is one antiderivative of $f$ on an interval, then every antiderivative there is $F(x) + C$ for some constant $C$. This is the corollary of the mean value theorem that two functions with equal derivatives differ by a constant.

Indefinite integral notation. The general antiderivative is written

$$ \int f(x)\,dx = F(x) + C $$

The symbol $dx$ names the variable; the constant $C$ is part of the answer, not decoration.

Power rule reversed. For $n \neq -1$,

$$ \int x^{n}\,dx = \frac{x^{n+1}}{n+1} + C $$

Raise the exponent by one and divide by the new exponent. Differentiating the right side returns $x^n$, which is the check.

The exceptional power. The rule above breaks at $n = -1$ because the new exponent would be zero. That case is covered by the logarithm:

$$ \int \frac{1}{x}\,dx = \ln|x| + C $$

The absolute value lets the formula serve negative inputs as well as positive ones.

Sums and constant multiples. Antidifferentiation respects sums and constant factors:

$$ \int \big[af(x) + bg(x)\big]\,dx = a\int f(x)\,dx + b\int g(x)\,dx $$

Trigonometric antiderivatives. Read the derivative rules backward, minding the signs:

$$ \int \cos x\,dx = \sin x + C, \qquad \int \sin x\,dx = -\cos x + C $$

$$ \int \sec^{2} x\,dx = \tan x + C, \qquad \int \sec x\tan x\,dx = \sec x + C $$

Exponential antiderivatives. From the derivative of the exponential,

$$ \int e^{x}\,dx = e^{x} + C, \qquad \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C \ \ (k \neq 0) $$

Initial value problem. A differential equation together with a condition such as $y(x_0) = y_0$. Antidifferentiate to get the general solution, then substitute the condition and solve for $C$ to get the one solution that fits.

Worked examples

Powers and roots

Evaluate $\displaystyle\int \left(3x^{2} - \frac{4}{x^{3}} + \sqrt{x}\right)dx$.

Rewrite every term as a power before touching it: $-4/x^3$ is $-4x^{-3}$, and $\sqrt{x}$ is $x^{1/2}$. Now the power rule applies term by term. The exponents become $3$, $-2$ and $3/2$:

$$ \int 3x^{2}\,dx = x^{3}, \qquad \int -4x^{-3}\,dx = -4\cdot\frac{x^{-2}}{-2} = 2x^{-2}, \qquad \int x^{1/2}\,dx = \frac{2}{3}x^{3/2} $$

So the answer is

$$ x^{3} + \frac{2}{x^{2}} + \frac{2}{3}x^{3/2} + C $$

Check by differentiating: the first term gives $3x^2$; the second, written as $2x^{-2}$, gives $-4x^{-3}$, which is $-4/x^3$; the third gives $\frac{2}{3}\cdot\frac{3}{2}x^{1/2} = \sqrt{x}$. All three match.

Trigonometric and exponential terms

Evaluate $\displaystyle\int \left(2e^{x} + 5\sec^{2}x - \frac{3}{x}\right)dx$.

Each term is on the list, so take them one at a time and keep the constant factors outside:

$$ 2e^{x} + 5\tan x - 3\ln|x| + C $$

Differentiate to check: $2e^x$ returns $2e^x$; $5\tan x$ returns $5\sec^2 x$; $-3\ln|x|$ returns $-3/x$. The most common slip here is a sign on a sine or cosine, so note that neither appeared, and that had the integrand contained $\sin x$ the answer would carry a minus sign in front of the cosine.

Recovering a function from its derivative

Solve the initial value problem $f'(x) = 6x - 4$ with $f(1) = 2$.

The general antiderivative is

$$ f(x) = 3x^{2} - 4x + C $$

Now use the condition. Substituting $x = 1$ gives $3 - 4 + C = 2$, so $C = 3$ and

$$ f(x) = 3x^{2} - 4x + 3 $$

Check both requirements: differentiating gives $6x - 4$, and $f(1) = 3 - 4 + 3 = 2$. A general antiderivative that satisfies the derivative condition but not the initial condition is only half an answer.

Two integrations for a trajectory

A ball is thrown upward from a height of $6$ ft with an initial speed of $48$ ft/s. Taking the acceleration due to gravity as $-32$ ft/s$^2$, find the height function and the maximum height.

Start from $a(t) = -32$. Antidifferentiating once gives the velocity:

$$ v(t) = -32t + C_1 $$

The initial speed fixes the constant: $v(0) = 48$ forces $C_1 = 48$, so $v(t) = -32t + 48$. Antidifferentiating again gives the height:

$$ s(t) = -16t^{2} + 48t + C_2 $$

and $s(0) = 6$ forces $C_2 = 6$, so $s(t) = -16t^2 + 48t + 6$.

The ball is at its highest point when the velocity is zero, that is when $-32t + 48 = 0$, so $t = 1.5$ s. Then

$$ s(1.5) = -16(2.25) + 48(1.5) + 6 = -36 + 72 + 6 = 42 $$

The maximum height is $42$ ft, reached $1.5$ s after release. Check the structure: $s'(t) = -32t + 48$ matches the velocity, and $s''(t) = -32$ matches the given acceleration, so both antidifferentiations were consistent.

Practice

Start with the reversal itself. These ask for general antiderivatives of powers, roots and polynomials, and for the solution of initial value problems where a data point pins down the constant.

Then the standard function list. These drill the indefinite integrals of trigonometric and exponential integrands, where most of the work is recognizing which derivative rule you are reading backward.

Quiz

Five items on general antiderivatives, the constant of integration, trigonometric and exponential rules, and initial value problems.