Lesson Part 1
So we've talked about differential calculus. Now we're going to enter the other section of calculus which is called integral calculus. So just as slopes of tangent lines and derivatives were fundamental for differential calculus, now we have other fundamental concepts which are going to be areas and antiderivatives.
In This Unit
- We will consider integral calculus.
- We will study fundamental ideas such as areas under curves and antiderivatives.
In This Module
- We will look at the idea of an antiderivative.
“Undoing” Differentiation
So if we remember all of our differentiation rules, we might remember the power rule. The power rule tells us how to differentiate powers of \(x\).
For example, we know that the derivative of \(x^2\) is \(2x\) by rolling down the constant and then reducing the power by \(1\). That is,
\[\dfrac{d}{dx} \Big ( x^2 \Big ) = 2x\]
The functions \(x^2\) and \(2x\) are connected by the process of differentiation.
We know that we call \(2x\) the derivative of \(x^2\).
We call \(x^2\), not too surprisingly, an antiderivative of \(2x\).
Finding an antiderivative of a function can be thought of as somewhat “undoing” differentiation.
In general, we're going to ask the following question:
Given a function \(f\), can we find another function capital \(F\) such that the derivative of capital \(F\) is our original function; i.e., such that \(F'=f\)?
So really, when finding the antiderivative of \(f(x)=2x\), we are asking: “What function did we differentiate to get \(2x\)?” And the answer to that question would be \(x^2\).
We call capital \(F\) an antiderivative of lowercase \(f\).
So the word we use here is “an” and the word “an” is very important because, while using our differentiation rules, we've seen there's only one derivative for any function. That's not going to be the case for antiderivatives.
That is, while any function, lowercase \(f\), has only one derivative, lowercase \(f\) will have more than one antiderivative.
In fact, if we can find one antiderivative of lowercase \(f\), then we can find infinitely many!
Examples
So now we'll move on to example 1 to start talking about antiderivatives.
Example 1
Check that \(\dfrac{1}{3}x^3\) and \(\dfrac{1}{3}x^3+2\) are both antiderivatives of the same function: \(f(x)=x^2\).
So what we're asking here is, “when we differentiate those two functions, do we get \(x^2\) as an answer?”
Solution
Now we're just talking about differentiation, so we have the power rule to differentiate polynomials.
So for the function \(\dfrac{1}{3}x^3\), if we differentiate that, we pull out the constant, \(\dfrac{1}{3}\), and we know the derivative of \(x^3\) is just \(3x\) squared by rolling down the constant and then taking the power down by \(1\). So with the \(\dfrac{1}{3}\) cancelling with the \(3\), we get down to our function, \(x^2\). That is,
\(\dfrac{d}{dx}\left ( \dfrac{1}{3}x^3\right )\) \(=\dfrac{1}{3}\left( \dfrac{d}{dx}\Big (x^3 \Big)\right)\) \(=\dfrac{1}{3} \left ( 3x^2\right) = x^2\)
So then the answer is yes, \(\dfrac{1}{3}x^3\) is indeed an antiderivative of \(f(x)=x^2\).
Now we check our second function which is just our original function with a constant \(2\) added. So if we differentiate this new function, we get \(\dfrac{1}{3}x^3 + 2\), all differentiated. Again, pulling out constants and evaluating, we get all the way down to \(\dfrac{1}{3} \left ( 3x^2\right)\), plus the derivative of \(2\), which is of course \(0\) because it's a constant, and we get down to \(x^2\). That is,
\(\dfrac{d}{dx}\left ( \dfrac{1}{3}x^3+2\right )\) \(= \dfrac{1}{3}\left( \dfrac{d}{dx}\Big (x^3 \Big) \right)+\dfrac{d}{dx}\Big(2\Big) \) \(=\dfrac{1}{3} \left ( 3x^2\right) + 0 \) \( = x^2\)
So again, the answer is yes in this case. Both of our functions are actually antiderivatives of \(f(x)=x^2\).
So now, one thing you might notice is that the constant \(2\) wasn't really special here. All that mattered was that \(2\) was a constant, and so the derivative of \(2\) was \(0\). We could replace \(2\) with any constant, \(c\), and the final answer would not change.
So this gives us a hint that when we're searching for antiderivatives, there's not just one or two. In fact, by changing the constant, we can come up with infinitely many antiderivatives.
Example 2
Let's move on to example 2, which is going to get a little bit harder. So now we're not going to check whether something's an antiderivative, we're going to actually try and find one from scratch.
Consider the function \(f(x)=x^5\). Find an antiderivative of \(f\).
Now again, remember “an” here, because we're not finding all of them. We're just finding one.
Solution
So the question we're asking is, “what function did we differentiate to get \(x^5\)?”
We know that the power rule for differentiating \(x^n\) reduces the power by \(1\). So likely, we're looking for something of the form \(x^6\). So when we differentiate, we lose \(1\) in the power. We have that the derivative of \(x^6\) is equal to \(6x^5\) by the power rule.
\[\dfrac{d}{dx}\Big ( x^6\Big )=6x^5\]
So it's almost the function we want, but it has this extra factor of \(6\) that we don't want. To fix this problem, we're just going to include an appropriate constant of \(\frac{1}{6}\) to cancel out with our \(6\). So we check the derivative of \(\frac{1}{6} x^6\) is indeed \(x^5\) when we get the cancellation of the \(\frac{1}{6}\) with the \(6\).
\[ \dfrac{d}{dx}\left ( \dfrac{1}{6} x^6 \right ) = \dfrac{1}{6}\left ( 6x^5 \right )= x^5\]
And so, captial \(F(x)=\dfrac{1}{6}x^6\) is an antiderivative of \(f(x)=x^5\).
So now we'll move on to another similar example
Example 3
Find an antiderivative of \(f(x)=2x^3\).
Solution
So again, we need to think of a function whose derivative is \(2x^3\).
By the power rule, we want to start with a function that has a larger exponent. In particular, we want to start with something of the form \(x^4\).
\[\dfrac{d}{dx} \left ( x^4 \right ) = 4x^3\]
So we're in the right ballpark here. Unfortunately, what we spat out is exactly \(2\) times the function that we want. So with a small adjustment, we can include \(\frac{1}{2}\) as a constant, and we check that our new function is indeed an antiderivative.
\[\dfrac{d}{dx}\left ( \dfrac{1}{2}x^4\right )=\dfrac{1}{2}\left ( 4x^3\right )=2x^3\]
Therefore, \(F(x)=\dfrac{1}{2}x^4\) is an antiderivative of \(f(x)=2x^3\).
You might notice, after doing a few examples, while it takes some effort to find an antiderivative of \(f\), it is easy to check if your answer is correct, because checking is just doing differentiation! So make sure you check your answers.
Also, hopefully, you noticed a pattern when attempting to find antiderivatives of powers of \(x\). If \(f(x)\) was a constant times \(x^5\), then the antiderivative we found was a constant times \(x^6\). If \(f(x)\) was a constant times \(x^2\), then the antiderivative we found was a constant times \(x^3\).
In general, if \(f(x)\) involves \(x^n\) for a positive integer \(n\), then any antiderivative will involve \(x^{n+1}\), one more than the power we started with. Let's look at the note shown here.
Note: If \(f(x) = x^n\), then one antiderivative of \(f(x)\) is captial \(F(x) = \dfrac{1}{n+1}x^{n+1}\).
Using the power rule, we can verify this fact. We differentiate capital \(F(x)\) and we'll find that we get \(f(x)\). This formula suggests that there is a more general or even systematic approach to finding antiderivatives. We can actually come up with a formula. This is in contrast to the guess and test approach that we used in this lesson. In the student exercises, you will be asked to explore this formula for more general exponents.
Lesson Part 2
So now we're going to start talking about finding all of the antiderivatives of a given function. So it's actually going to be an entire family of functions.
Family of Functions
As discussed in example 1, there are infinitely many antiderivatives of the function \(x^2\).
In particular, it can be shown that any function of the form \(\frac{1}{3}x^3+c\), where \(c\) is constant, is an antiderivative of \(x^2\), and we can check this. So does every antiderivative of \(x^2\) have to look like this, or are there possibly some other ones?
So now we have to go back and think about what we know about differentiation. We know the following properties:
- If two functions \(F\) and \(G\) satisfy \(F'=G'\), then \(F\) and \(G\) must differ by at most a constant. So either they're the same function, or maybe \(F = G + 2\).
- If \(F\) and \(G\) are two antiderivatives of a common function lowercase \(f\), then \(F'=\text{lowercase }f=G'\) and so they have the same derivative. That tells us that \(F=G+c\) for some constant \(c\).
So in fact, every antiderivative of \(x^2\) has to be of the form \(\frac{1}{3}x^3+c\).
Examples
So let's move on to an example dealing with finding all of the antiderivatives.
Example 4
Find all antiderivatives of the function \(f(x)=3x^4\).
Solution
So what we just talked about is, well, if we can find one antiderivative \(F\), then we can find them all by just adding the constant. Let's try and find a single antiderivative for our function.
Using the same strategy as before, we have something which looks like \(x^4\), so, with the power rule, we're looking for something of the form \(x^{4+1} = x^5\). We differentiate:
\[\dfrac{d}{dx}\left ( x^5 \right ) = 5x^4\]
So we need to make a similar adjustment with a constant. Well, we have a \(5\) there we don't want, and we need to introduce a \(3\). So we introduce the \(3\) in the numerator, and we introduce a factor of \(\dfrac{1}{5}\) to cancel out the \(5\). So we're left with the function \(\dfrac{3}{5}x^5\). That is,
\[x^5 \implies 3x^5 \implies \dfrac{1}{5}\left ( 3x^5 \right ) \implies \dfrac{3}{5}x^5\]
And indeed, that differentiates to give
\[\dfrac{d}{dx}\left ( \dfrac{3}{5}x^5\right ) = 3x^4\]
Thus, any antiderivative must be of the form \(F(x)=\frac{3}{5}x^5\) is an antiderivative of \(f(x)=3x^4\).
Any other antiderivative of \(f\) must differ from \(\frac{3}{5}x^5\) by only a constant.
Thus, any antiderivative must be of the form \(F(x)=\frac{3}{5}x^5+c\), for some constant \(c\).
Quickly checking to confirm that all of those functions indeed have the right derivative, we see
\[\dfrac{d}{dx} \left ( \dfrac{3}{5}x^5+c\right ) = \dfrac{3}{5}\left ( \dfrac{d}{dx}\Big ( x^5 \Big )\right) +\dfrac{d}{dx}\Big( c \Big) = \dfrac{3}{5}\left ( 5x^4 \right) +0 = 3x^4\]
for all constants \(c\).
Example 5
Let's move on from powers for a second and talk about a different function.
Find all antiderivatives of the function \(f(x)=3e^x\).
Solution
Recall that \(\dfrac{d}{dx}\Big (e^x \Big )=e^x\), so an antiderivative of \(e^x\) is \(e^x\).
Well, the constant doesn't change much there, so, similarly,
\[\dfrac{d}{dx} \Big (3e^x \Big ) = 3e^x\]
So \(3e^x\) is an antiderivative of \(3e^x\).
To find the family of all antiderivatives, we just add our constant.
It follows that every antiderivative of \(f(x)=3e^x\) is of the form
\[F(x) = 3e^x +c\]
for some constant c.
Similarly, we can find antiderivatives for things like the sine function, the cosine function, and basically any derivative rule we have is going to help us find antiderivatives of functions. But we'll leave most of that discussion for another time.
| Function |
Antiderivative |
| \(1\) |
\(x\) |
| \(x\) |
\(\dfrac{1}{2}x^{2}\) |
| \(e^{x}\) |
\(e^{x}\) |
| \(\sin(x)\) |
\(-\cos(x)\) |
| \(\cos(x)\) |
\(\sin(x)\) |
Lesson Part 3
The last thing we want to talk about in this lesson is dealing with negative powers.
Examples
Example 6
Find an antiderivative of \(f(x)=\dfrac{2}{x^2}\).
Solution
Notice that \(f(x)=\dfrac{2}{x^2} = 2x^{-2}\). Here, the exponent is negative.
We've looked at only polynomials up till now, but the power rule includes negative powers as well, so the method should be the same as even if we're including negative powers of \(x\). In fact, although the power is negative, the method does stay the same.
So again, we're looking for “what function do we differentiate to get \(2x^{-2}\)?”
By the power rule, we can see
\[\dfrac{d}{dx}\left ( \dfrac{1}{x} \right ) =\dfrac{d}{dx}\left ( x^{-1}\right )= -x^{-2}\]
So we're in the right family here. We're just off by a constant. We have a negative in there and we're missing a \(2\).
So with a correction, we can include a constant, \(-2\), and then check our answer. So if we differentiate, we indeed get our desired function:
\[\dfrac{d}{dx}\left ( -\dfrac{2}{x}\right) = \dfrac{d}{dx} \Big (-2x^{-1}\Big ) = -2 \left (-x^{-2} \right )= \dfrac{2}{x^2}\]
Therefore, \(F(x)=-\dfrac{2}{x}\) is an antiderivative of \(f(x)=\dfrac{2}{x^2}\).
Notice this is the first time we've seen a function that's not defined everywhere. So of course, because \(f(x)=\dfrac{2}{x^2}\) has an \(x\) in the denominator, it's not defined for \(x = 0\).
But we'll notice that the antiderivative of our function has the same property. So our functions here have the same domain. In other words, the antiderivative \(F\) is defined over the whole domain of \(f\) (they're defined everywhere but \(x = 0\)).
Example 7
Now we're going to move on to a more challenging example to finish off. You might have noticed we've talked about powers of \(x\) for the most part, but there's one special power of \(x\) that takes a little more work.
Find an antiderivative of \(f(x)=\dfrac{1}{x}\).
Solution
So we go to our usual attempt and we look at the power rule. But pretty quickly, we notice that the power rule fails since there is no power of \(x\) whose derivative is \(x^{-1}\). The only chance would be something of the form \(x^0\), but of course, that's a constant, and so the derivative is \(0\) and not what we want. So for the first time, we have to look for a new method.
Luckily, we know a function whose derivative is \(\dfrac{1}{x}\). Recall that \(\dfrac{d}{dx} \Big ( \ln (x) \Big ) = \dfrac{1}{x}\).
So the only problem we see here is that, of course, the function \(\dfrac{1}{x}\) is defined for all \(x\) except for \(x = 0\), but we know that \(\ln (x)\) is actually defined only for \(x \gt 0\). So we're going to talk about \(\dfrac{1}{x}\) being the derivative of \(\ln\). We'd at least like them to be defined on the same domain.
Therefore, the function \(\ln(x)\) is not the best choice for an antiderivative of \(\dfrac{1}{x}\) as they have different domains.
Can we find a function capital \(F(x)\), defined, at least, for all \(x\neq 0\), satisfying capital \(F'=f\)?
We are looking for a function that is like \(\ln (x)\), but that is defined for all non-zero \(x\) values.
Let's consider the missing interval \(x \lt 0\). Notice that the function \(\ln(-x)\) is defined for all \(x \lt 0\), and it is certainly related to \(\ln (x)\).
Let's examine the derivative of \(\ln(-x)\). By the chain rule, the derivative is \(\dfrac{1}{(-x)}\) times the derivative of negative \(x\), which is \(-1\). This simplifies to \(\dfrac{1}{x}\). That is,
\[\dfrac{d}{dx}\Big(\ln(-x)\Big) = \dfrac{1}{(-x)}(-1) = \dfrac{1}{x}\]
So what do we have here? We have a second function, \(\ln(-x)\), that is also an antiderivative of \(\dfrac{1}{x}\). It's no better than \(\ln (x)\), however, as it's only defined for half of the real numbers, essentially, for \(x \lt 0\), while \(\dfrac{1}{x}\) is still defined for all non-zero \(x\).
Therefore, the function \(\ln(-x)\) is also an antiderivative of \(\dfrac{1}{x}\), but with domain \(x \lt 0\).
So it looks like we need to somehow glue these functions together to get the full picture of an antiderivative for \(\dfrac{1}{x}\).
Let's define the following piecewise function with domain \(x \neq 0\), the same domain as \(\dfrac{1}{x}\):
\[f(x) = \displaystyle\left\{\begin{array}{ll}\ln(x) & \text{if } x \gt 0 \\ \ln(-x) & \text{if } x \lt 0\end{array}\right.\]
Then for all non-zero \(x\), \(f(x)\) is defined, and \(f'(x)=\dfrac{1}{x}\) from our earlier work.
Observe that capital \(F(x) = \ln\left({\lvert x \rvert}\right)\), since \(\lvert x \rvert = x\) when \(x \gt 0\) and \(\lvert x \rvert = -x\) when \(x \lt 0\).
We have shown that capital \(F(x) = \ln\left({\lvert x \rvert}\right)\) is an antiderivative of \(f(x) = \dfrac{1}{x}\) with the same domain as \(\dfrac{1}{x}\).
Conclusion
So I guess the moral of the story here is that finding an antiderivative is generally a bit harder than finding a derivative. But it's easy to check your work because, while we have to take, sometimes, a wild stab at the antiderivative, we can always check pretty quickly whether we're right. So it's always a good idea to check.
It is easy to check if an antiderivative is correct \(-\) just differentiate.
So now we're going to leave the topic of antiderivatives for a while, and discuss the fundamental concept of integral calculus, which is the notion of area. But our antiderivatives will come back later in a big way to solve some really tough problems for us.
Quiz
See the quiz in the side navigation.