不定积分是独一无二的吗?

In other words, here's the question:

Can a function have more than one anti-derivative?

So, a function can have more than one anti-derivative. Here are some examples of anti-derivatives of $f(x) = x$, which you might have found:

$\bullet$ $\frac{x^2}{2}$

$\bullet$ $\frac{x^2}{2} + 1$

$\bullet$ $\frac{x^2}{2} - 1$

$\bullet$ $\frac{x^2}{2} + 100$

So, we can add any constant (that is, any number that doesn't depend on $x$) to $\frac{x^2}{2}$ and get another anti-derivative.

Because of this, if we're asked for the anti-derivative of $f(x) = x$, the best answer is this:

The anti-derivative of $f(x) = x$ is $\frac{x^2}{2} + \mathcal{C}$ for any constant $\mathcal{C}$.

Let's try to turn this into a general rule:

If $F(x)$ is an anti-derivative of $f(x)$, then for any constant $c$, $F(x) + \mathcal{C}$ is also an anti-derivative of $f(x)$.

Basically, this rule says that if we can find one anti-derivative of a function, then we can add any constant and find many more anti-derivatives of that function.

Your proof should look something like this:

Since the derivative of a sum is the sum of the derivatives,

$$\frac{d}{dx}[F(x) + \mathcal{C}] = \frac{d}{dx}[F(x)] + \frac{d}{dx}[\mathcal{C}]$$

The derivative of $F(x)$ is $f(x)$ (since $F(x)$ is an anti-derivative of $f(x)$), and the derivative of $\mathcal{C}$ is $0,$ so the derivative of $F(x) + \mathcal{C}$ is $f(x)$. Therefore, $F(x) + \mathcal{C}$ is an anti-derivative of $f(x)$.

We have one more question left to ask, and hopefully it's on your mind already:

Does finding one anti-derivative and adding constants give you all of the anti-derivatives of a function?

Here's another way to frame the same question:

Do any two anti-derivatives of a function always differ by a constant?

Let's try to answer the second question.

Let $f(x)$ be a function, and suppose $F(x)$ and $G(x)$ are two anti-derivatives of $f(x)$.

Your proof should look something like this:

Since the derivative of a difference is the difference of the derivatives,

$$\frac{d}{dx}[F(x) - G(x)] = F'(x) - G'(x) = f(x) - f(x) = 0$$

Therefore, $F(x) - G(x)$ is a constant. (See here for an explanation of this last step.)

So, to find all anti-derivatives of a function, we can find one anti-derivative and add any constant $\mathcal{C}$.

Are anti-derivatives unique? In one sense, no; functions never have just one. But in another sense, yes; if we think of functions that differ by a constant as essentially the same function, then we've just proved that a function can essentially only have one anti-derivative.

To put this precisely, we say that anti-derivatives are unique up to adding a constant.

Look at the answers to the problems on this worksheet. Then take the antiderivatives of the answers. Did you get the same antiderivative as the problem? If not, what's the difference? Here's some food for thought on Black Friday!