Antiderivatives

Running differentiation backwards

Differentiation takes a function and gives its slope. An antiderivative goes the other way: given a function $f$, find a function $F$ whose slope is $f$. In symbols, $F'(x) = f(x)$.

If $f$ is a speed in km per hour, an antiderivative is a position in km, since the slope of position is speed. Lesson 2.5.3 uses that link to find areas.

Guess, check, adjust

To find an antiderivative of $x^2$, guess $x^3$: differentiating lowers a power by one, so the answer must have power 3. Then differentiate the guess and compare.

The + C: a family of antiderivatives

$\frac{x^3}{3}$ is not the only antiderivative of $x^2$. So are $\frac{x^3}{3} + 5$ and $\frac{x^3}{3} - 2$, because an added constant vanishes when you differentiate (lesson 2.3.1).

On an interval, these are all of them: two functions with the same derivative differ by a constant. The indefinite integral collects the whole family under one symbol.

On a graph, the family is one curve copied up and down. At any $x$ the copies have the same slope, so their tangent lines there are parallel.

One value picks one curve

A single known value of $F$ fixes $C$. Say $F'(x) = 2x$ and $F(1) = 3$. Then $F(x) = x^2 + C$, and putting in $x = 1$ gives $1 + C = 3$, so $C = 2$ and $F(x) = x^2 + 2$.

If $F$ is a position and $F'$ its speed, the speed alone cannot tell you where the car started. One known position settles it.

Powers

Read the power rule of lesson 2.3.1 backwards. Differentiating lowers a power by one and multiplies by the old power; undoing it raises the power by one and divides by the new power.

Write roots and reciprocals as powers first (lesson 1.4.1). Each line of the table leaves out the $+ C$.

Check the fourth line by differentiating: $\frac{d}{dx}\left(-x^{-1}\right) = (-1)(-1)\,x^{-2} = x^{-2}$. An antiderivative keeps the gaps in its domain. $\frac{1}{x^2}$ and $-\frac{1}{x}$ have no value at 0, so the answer holds for $x > 0$ and for $x < 0$ separately.

Constant multiples and sums

So $\int (6x^2 - 4x + 5)\,dx = 2x^3 - 2x^2 + 5x + C$. Differentiating gives back $6x^2 - 4x + 5$.

There is no product rule backwards. $\int x \cdot x\,dx = \frac{x^3}{3}$, but multiplying the separate antiderivatives gives $\frac{x^2}{2} \cdot \frac{x^2}{2} = \frac{x^4}{4}$, which is wrong. Multiply out first when you can.

For example, $\int (x + 1)^2\,dx = \int (x^2 + 2x + 1)\,dx = \frac{x^3}{3} + x^2 + x + C$. The chain rule (lesson 2.3.3) shows that $\frac{(x + 1)^3}{3}$ works too. Multiplied out, it is $\frac{x^3}{3} + x^2 + x + \frac{1}{3}$: the two answers differ by a constant, which the $+ C$ absorbs.

The exception: $\frac{1}{x}$

For $n = -1$ the power rule would give $\frac{x^0}{0}$, a division by zero. The antiderivative of $\frac{1}{x}$ is a logarithm instead.

Lesson 2.3.4 showed that $\ln x$ has slope $\frac{1}{x}$ for $x > 0$. It also showed that $\ln|x|$, the same curve mirrored in the $y$-axis, has slope $\frac{1}{x}$ for $x < 0$ too. Read backwards, that is the antiderivative on both sides.

For example, at $x = -2$ the slope of $\ln|x|$ is $-0.5$, which is $\frac{1}{-2}$. Constants come out as usual: $\int \frac{3}{x}\,dx = 3\ln|x| + C$.

Exponentials

$e^x$ is its own derivative, so it is its own antiderivative. For $e^{kx}$, differentiating brings out a factor $k$ (lesson 2.3.4), so the antiderivative divides by $k$.

So $\int e^{3x}\,dx = \frac{e^{3x}}{3} + C$. And $\int e^{-x/2}\,dx = -2e^{-x/2} + C$, because dividing by $k = -\frac{1}{2}$ multiplies by $-2$. Case study 2.5.5 integrates a discount factor $e^{-rt}$ in exactly this way.

Other bases follow from $\frac{d}{dx}b^x = b^x \ln b$ (lesson 2.3.4): $\int b^x\,dx = \frac{b^x}{\ln b} + C$. So $\int 2^x\,dx = \frac{2^x}{\ln 2} + C$. Lesson 2.6.1 extends the trick for $e^{kx}$ into a general method, substitution.

Sine and cosine

Lesson 2.3.5 showed that $\sin x$ has slope $\cos x$, and $\cos x$ has slope $-\sin x$. Read backwards, $\sin x$ is an antiderivative of $\cos x$, and $-\cos x$ is an antiderivative of $\sin x$.

The minus sign goes with the sine. $\cos x$ is an antiderivative of $-\sin x$, not of $\sin x$.

With a factor inside, divide by it, as for $e^{kx}$: $\int \cos(3x)\,dx = \frac{\sin(3x)}{3} + C$. Lesson 2.5.3 turns antiderivatives into exact areas. For $\sin x$ on $[0, \pi]$ that gives $-\cos\pi + \cos 0 = 2$, the area of the hump that lesson 2.5.1 estimated.

When there is no formula

Every function made of powers, exponentials and logs can be differentiated by rules. Antiderivatives are harder: some exist but cannot be written with those functions at all.

$e^{-x^2}$ is the standard example. It has antiderivatives (lesson 2.5.3 builds one from areas), but no combination of powers, roots, exponentials and logs has derivative $e^{-x^2}$. SymPy answers with a new function, erf, which is itself defined by an integral of $e^{-x^2}$.

Lesson 2.6.3 meets its close relative $e^{-x^2/2}$, the bell curve, whose areas are found numerically. Other integrands need a method rather than a rule: substitution (lesson 2.6.1) and integration by parts (lesson 2.6.2).

Common mistakes

1. Divide by the new power. $\int x^3\,dx = \frac{x^4}{4}$, not $\frac{x^4}{3}$, and not $3x^2$, which is the derivative.
2. Keep the $+ C$. An indefinite integral is a whole family. When a value of $F$ is given, use it to find $C$.
3. $n = -1$ is the exception. $\int \frac{1}{x}\,dx = \ln|x| + C$, not $\frac{x^0}{0}$.
4. $\frac{1}{x^2}$ is not a log case. It is $x^{-2}$, so the power rule gives $-\frac{1}{x}$.
5. Divide by $k$, do not multiply. $\int e^{3x}\,dx = \frac{e^{3x}}{3} + C$, not $3e^{3x}$.
6. There is no product rule backwards. $\int x \cdot x\,dx$ is $\frac{x^3}{3}$, not $\frac{x^2}{2} \cdot \frac{x^2}{2}$.
7. Mind the sign with sine. $\int \sin x\,dx = -\cos x + C$, not $\cos x + C$.
8. Two answers can both be right. $\frac{x^3}{3} + x^2 + x$ and $\frac{(x + 1)^3}{3}$ differ by a constant, so both are antiderivatives of $(x + 1)^2$.