Dictum 1.1.1. Calculus is the science of functions.
In more detail, calculus investigates and analyzes properties of functions using three fundamental tools: the limit, the derivative, and the integral.
Section 1.1 Antiderivatives
What is calculus? Below you find a decent, if terse answer.
Presumably you are already on intimate terms with the limit and the derivative. In this course we take up the integral in earnest. We begin somewhat indirectly, introducing first the antiderivative and its accompanying indefinite integral notation. Both of these concepts have everything to do with the derivative, and nothing directly to do with the actual definition of the integral! However, as we will see with the
[provisional cross-reference: th_ftc], these concepts will play an essential role in computing integrals.Definition 1.1.2. Antiderivative.
Let \(f\) be a function defined on an interval \(I\text{.}\) An antiderivative of \(f\) on \(I\) is a function \(F\) satisfying \(F'(x)=f(x)\) for all \(x\in I\text{.}\)
Example 1.1.3. Basic antiderivative computations.
Find an antiderivative for the given function \(f\) on the given interval \(I\text{.}\)
- \(f(x)=x^7\text{,}\) \(I=\R\)
- \(f(x)=\frac{1}{\sqrt{x}}\text{,}\) \(I=(0,\infty)\)
- \(f(x)=2\sin x-x^{2/3}\text{,}\) \(I=\R\)
Solution.
We find antiderivatives for each example “by inspection”. To check that our answer is correct, we simply verify that for each candidate antiderivative \(F\) we have \(F'(x)=f(x)\) for all \(x\) in the given interval \(I\text{.}\) We leave this to you.
- \(F(x)=\frac{1}{8}x^8\) is an antiderivative of \(f(x)=x^7\) on \(I=\R\text{.}\)
-
\(F(x)=2\sqrt{x}\) is an antiderivative of \(f(x)=\frac{1}{\sqrt{x}}\) on \(I=(0,\infty)\text{.}\)To verify that \(F'(x)=f(x)\) here it is essential to translate the given radical expressions to power expressions so that we can make use of the power rule. That is, we have \(F(x)=2x^{1/2}\) and \(f(x)=\frac{1}{x^{1/2}}=x^{-1/2}\text{,}\) whence \(F'(x)=x^{-1/2}=f(x)\) for all \(x\in (0,\infty)\text{.}\)
- \(F(x)=-2\cos x-\frac{3}{5}x^{5/3}\) is an antiderivative of \(f(x)=2\sin x-x^{2/3}\) on \(I=\R\)
Example 1.1.4. Antiderivatives depend on intervals.
Let \(f(x)=\abs{x}\text{.}\) Show that \(F(x)=\frac{1}{2}x^2\) is an antiderivative of \(f\) on \(I=[0,1]\text{,}\) but not on \(J=[-1,0]\text{.}\) Find an antiderivative of \(f\) on \(J\text{.}\)
Solution.
We have \(F'(x)=x\) for all \(x\in \R\text{.}\) Since by definition we have
\begin{equation*}
f(x)=\abs{x}=\begin{cases} x \amp\text{if } x\geq 0 \\ -x \amp x \leq 0 \end{cases}\text{,}
\end{equation*}
we see that \(F'(x)=f(x)=x\) for all \(x\in [0,1]\text{.}\) Thus \(F\) is an antiderivative of \(f\) on \(I\text{.}\)
By the same token, since \(F'(x)=x\) and \(f(x)=-x\) on \(J=[-1,0]\text{,}\) we see that \(F'(x)\ne f(x)\) for all \(x\in J\text{,}\) and hence that \(F\) is not an antiderivative of \(f\) on \(J\text{.}\)
Lastly, from our analysis above we now see that \(G(x)=-\frac{1}{2}x^2\) is an antiderivative of \(f\) on \(J=[-1,0]\text{,}\) since on this interval we have \(G'(x)=f(x)=-x\text{.}\)
Procedure 1.1.5. Computing antiderivatives by inspection.
Given a function \(f\) and interval \(I\text{,}\) to compute (with justification) an antiderivative \(F\) of \(f\) by inspection, proceed as follows.
- Provide a candidate antiderivative \(F\text{.}\)
- Verify that \(F\) is an antiderivative by showing that \(F'(x)=f(x)\) for all \(x\in I\text{.}\)
This is very much a “guess and check” procedure. Do not be shy in hazarding a guess for your candidate antiderivative \(F\text{;}\) even if your check proves the candidate to be incorrect, the very act of checking will often reveal what needs to be adjusted to produce an actual antiderivative.
Example 1.1.6. Less elementary antiderivative computations.
Find an antiderivative for the given function \(f\) on the given interval \(I\text{.}\)
- \(f(x)=\sec^2 5x\text{,}\) \(I=(-\pi/2, \pi/2)\)
- \(f(x)=x\cos(x^2)\text{,}\) \(I=\R\)
- \(f(x)=\cos(x^2)\text{,}\) \(I=\R\)
Solution.
Part of what makes these computations more difficult is that the chain rule inevitably arises when checking our candidate antiderivatives. Since the chain rule plays an important role here, make sure you indicate where it arises in your check.
- \(F(x)=\frac{1}{5}\tan 5x\) is an antiderative of \(f(x)=\sec^2 5x\) on \(I=(-\pi/2, \pi/2)\) since\begin{align*} \frac{d}{dx} \frac{1}{5}\tan 5x \amp = \frac{1}{5}\left( \frac{d}{dx}\tan 5x \right) \amp \text{(scalar rule)} \\ \amp = \frac{1}{5}\sec^2 5x \cdot (5x)' \amp \text{(chain rule)}\\ \amp = \frac{1}{5}(5\sec^2 5x)\\ \amp = \sec^2 5x\text{.} \end{align*}
- \(F(x)=\frac{1}{2}\sin(x^2)\) is an antiderivative of \(f(x)=x\cos(x^2)\) since\begin{align*} F'(x) \amp = \frac{1}{2}\left( \sin(x^2)\right)' \amp \text{(scalar rule)}\\ \amp = \frac{1}{2}\cos(x^2)\cdot (x^2)' \amp \text{(chain rule)}\\ \amp = \frac{1}{2}(2x\cos(x^2))\\ \amp = x\cos(x^2)\text{.} \end{align*}
- Hmmm. This is a curious one. Some typical educated guesses as to what \(F\) might be (e.g., \(F(x)=\sin(x^2)\) or \(F(x)=\frac{\sin(x^2)}{x^2}\)) all prove incorrect when carefully computing \(F'(x)\) using the chain and/or product rules. Two natural questions arise: (1) does \(f(x)=\cos(x^2)\) have an antiderivative?; (2) if it does, how do we write it down? All will be revealed in good time!
Theorem 1.1.7. General antiderivative formula.
Assume \(F\) is an antiderivative of \(f\) on the interval \(I\text{.}\)
- Given any constant \(C\in \R\text{,}\) the function \(G(x)=F(x)+C\) is also an antiderivative of \(f\) on \(I\text{.}\)
- Conversely, if \(G\) is an antiderivative of \(f\) on \(I\text{,}\) then there is a constant \(C\in\R\) such that \(G(x)=F(x)+C\) for all \(x\in I\text{.}\)
Proof.
Let \(F\) be an antiderivative of \(f\) on the interval \(I\) and let \(C\in \R\) be any real constant. We have
\begin{align*}
\frac{d}{dx}\left(F(x)+C\right) \amp = \frac{d}{dx}F(x)+\frac{d}{dx}(C)\\
\amp =F'(x)+0\\
\amp = f(x)
\end{align*}
for all \(x\in I\text{.}\) Thus \(G(x)=F(x)+C\) is an antiderivative for any \(C\in \R\text{.}\)
Assume \(F\) and \(G\) are antiderivatives of \(f\) on the interval \(I\text{,}\) and define \(H(x)=F(x)-G(x)\text{.}\) We have
\begin{equation*}
H'(x)=F'(x)-G'(x)=f(x)-f(x)=0
\end{equation*}
for all \(x\in I\text{.}\) It follows (as a consequence of the mean value theorem) that \(H\) is a constant function on \(I\text{.}\) In other words, we have \(H(x)=C\) for all \(x\in I\text{.}\) Since \(H(x)=F(x)-G(x)\text{,}\) we conclude that
\begin{equation*}
G(x)=F(x)+H(x)=F(x)+C
\end{equation*}
for all \(x\in I\text{,}\) as desired.
Remark 1.1.8.
The two statements of Theorem 1.1.7 taken together are equivalent to the following: if \(F\) is an antiderivative of \(f\) on the interval \(I\text{,}\) then the set of all antiderivatives of \(f\) on \(I\) is given by
\begin{equation*}
\{G(x)\colon G(x)=F(x)+C \text{ for some } C\in \R\}\text{.}
\end{equation*}
In other words, once we find one antiderivative \(F\) of \(f\) on \(I\text{,}\) we obtain all other antiderivatives of \(f\) simply by adding an arbitrary constant \(C\) to \(F\text{.}\)
In particular, note that antiderivatives are not unique! There is no such thing as the antiderivative of a function; if \(f\) has an antiderivative \(F\text{,}\) then it has infinitely many antiderivatives \(F(x)+C\text{,}\) where \(C\) is any real constant.
Example 1.1.9. Leaking water tank.
At time \(t=0\) minutes a 100 gallon tank of water begins leaking. After \(t\) minutes, the rate at which the gallon leaks is
\begin{equation*}
r(t)=\frac{1}{\sqrt{2t+1}} \text{ gallons per minute}\text{.}
\end{equation*}
Find a formula for \(f(t)\text{,}\) the amount of water in the tank after \(t\) minutes.
Solution.
Recall that if \(Q=f(t)\text{,}\) then the function \(f'(t)\) outputs the instantaneous rate of change of \(Q\) with respect to \(t\) at the given time. Thus in our current setting \(f'(t)\) is the rate of change of the amount of water in the tank at time \(t\text{.}\) We are told that water leaks out of the tank at a rate given by \(r(t)\text{.}\) It follows that
\begin{equation*}
f'(x)=-r(t)=\frac{-1}{\sqrt{2t+1}}\text{.}
\end{equation*}
Put another way, \(f\) is an antiderivative of \(-r(t)=-1/\sqrt{2t+1}\text{.}\) An application of Procedure 1.1.5 yields \(F(t)=-\sqrt{2t+1}\) as an antiderivative of \(-r(t)\text{.}\) It follows from Theorem 1.1.7 that the general antiderivative is of the form \(F(x)+C=-\sqrt{2t+1}+C\text{.}\) Since \(f\) is also an antiderivative, we thus have
\begin{equation}
f(t)=-\sqrt{2t+1}+C\text{.}\tag{1.1.1}
\end{equation}
To determine what the constant \(C\) is, we use the initial condition \(f(0)=100\text{.}\) Using (1.1.1) this implies that
\begin{equation*}
100=f(0)=-\sqrt{1}+C=-1+C\text{.}
\end{equation*}
We conclude that \(C=101\) and thus \(f(t)=101-\sqrt{2t+1}\text{.}\)
Example 1.1.10. Initial value problem.
Consider the differential equation
\begin{equation}
f''(x)=-\frac{2}{3}\cos 2x+x\text{.}\tag{1.1.2}
\end{equation}
Solution.
- First, since \(f'(x)\) is an antiderivative of \(f''(x)=-\frac{2}{3}\cos 2x+x\text{,}\) using Procedure 1.1.5 and Theorem 1.1.7, we conclude that\begin{equation} f'(x)=-\frac{1}{3}\sin 2x+\frac{1}{2}x^2+C\tag{1.1.3} \end{equation}for some \(C\in \R\text{.}\) Next, since \(f(x)\) is an antiderivative of \(f'(x)=-\frac{1}{3}\sin 2x+\frac{1}{2}x^2+C\text{,}\) reasoning as above we have\begin{equation} f(x)=\frac{1}{6}\cos 2x+\frac{1}{6}x^3+Cx+D\tag{1.1.4} \end{equation}
- The given initial conditions \(f(0)=0\) and \(f'(0)=-1\) allows us to solve for the \(C\) and \(D\) in (1.1.4) as follows. First, using (1.1.3) we have\begin{equation*} -1=f'(0)=0+0+C\text{,} \end{equation*}\begin{equation*} 0=f(0)=\frac{1}{6}+0+0+D\text{,} \end{equation*}or \(D=-1/6\text{.}\) We conclude that\begin{equation*} f(x)=\frac{1}{6}\cos 2x+\frac{1}{6}x^3-x-\frac{1}{6}\text{.} \end{equation*}
Definition 1.1.11. Indefinite integral.
If \(F\) is an antiderivative of \(f\) on the interval \(I\text{,}\) then by Theorem 1.1.7 any other antiderivative is of the form \(G(x)=F(x)+C\) for some \(C\in \R\text{.}\) We say in this case that the expression \(F(x)+C \) is the general antiderivative and denote this using the indefinite integral notation
\begin{equation}
\int f\, dx=F(x)+C\text{.}\tag{1.1.5}
\end{equation}
The symbol \(\int\) is called the integral symbol, the function \(f\) is called the integrand of the integral, and \(x\) is called the variable of integration.
Remark 1.1.12.
It should be noted that the indefinite integral notation is riddled with ambiguity. In particular, whereas the interval of definition \(I\) plays an important role in the definition of antiderivative (as we saw in Example 1.1.4), it does not appear in the notation \(\int f\, dx\text{.}\) So how are we to understand a statement like (1.1.5)? We will take it to mean that on some interval (either implied or given) the function \(F\) is an antiderivative of \(f\text{,}\) and hence that the general antiderivative formula for \(f\) on this interval is given by \(G(x)=F(x)+C\text{.}\)
Example 1.1.13. Indefinite integral formulas.
Some of our previous example computations can be summarized by the following indefinite integral formulas.
\begin{align*}
\int x^7\, dx \amp = \frac{1}{8}x^8+C\\
\int \frac{1}{\sqrt{x}}\, dx \amp = 2\sqrt{x}+C\\
\int 2\sin x-x^{2/3}\, dx \amp = -2\cos x-\frac{3}{5}x^{5/3}+C\\
\int \sec^2 5x\, dx \amp = \frac{1}{5}\tan 5x +C\\
\int x\cos(x^2)\, dx \amp = \frac{1}{2}\sin (x^2)+C\text{.}
\end{align*}
Theorem 1.1.14. Antiderivative formulas.
The following antiderivative (or indefinite integral) formulas follow directly from a corresponding derivative formula. The \(k\) appearing in the formulas below is understood to be any nonzero constant.
\begin{align*}
\int 0\, dx\amp =C \amp \int x^r\, dx\amp =\frac{x^{r+1}}{r+1}+C, r\ne -1\\
\int \cos kx\, dx\amp =\frac{1}{k}\sin kx+C \amp \int \sin kx\, dx \amp =-\frac{1}{k}\cos kx+C \\
\int\sec^2kx\, dx\amp =\frac{1}{k}\tan kx+C \amp \int \csc^2 kx \, dx\amp =-\frac{1}{k}\cot kx+C \\
\int \sec kx\tan kx\, dx\amp =\frac{1}{k}\sec kx+C \amp \int \csc kx\cot kx\, dx\amp =-\frac{1}{k}\csc kx+C \text{.}
\end{align*}
Theorem 1.1.15. Antiderivative rules.
If \(F\) and \(G\) are antiderivatives of the functions \(f\) and \(g\text{,}\) respectively, on the interval \(I\text{,}\) then (1) \(cF\) is an antiderivative of \(cf\) for any \(c\in\R\text{,}\) (2) \(F+G\) is an antiderivative of \(f+g\text{,}\) and hence \(cF\pm dG\) is an antiderivative of \(cf\pm dg\) for any \(c,d\in\R\text{,}\)
A somewhat liberal usage of indefinite integral notation allows us to summarize these results as follows.
-
Scalar rule.\(\int cf\, dx = c\int f\, dx\)
-
Sum rule.\(\int f+g \, dx = \int f\, dx+\int g\, dx\)
-
Linear combination rule.\(\int cf\pm dg\, dx= c\int f\, dx \pm d\int g\, dx\)
.
