Antiderivatives

Section 1.1 Antiderivatives

What is calculus? Below you find a decent, if terse answer.

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.

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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.

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Definition 1.1.2. Antiderivative.

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Let

f

be a function defined on an interval

I .

An antiderivative of

f

I

is a function

F

satisfying

F^{'} (x) = f (x)

for all

x \in I .

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Example 1.1.3. Basic antiderivative computations.

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Find an antiderivative for the given function

f

on the given interval

I .

$f (x) = x^{7},$ $I = R$
$f (x) = \frac{1}{\sqrt{x}},$ $I = (0, \infty)$
$f (x) = 2 \sin x - x^{2 / 3},$ $I = R$

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 .

We leave this to you.

$F (x) = \frac{1}{8} x^{8}$ is an antiderivative of $f (x) = x^{7}$ on $I = R .$
$F (x) = 2 \sqrt{x}$ is an antiderivative of $f (x) = \frac{1}{\sqrt{x}}$ on $I = (0, \infty) .$

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 $F (x) = 2 x^{1 / 2}$ and $f (x) = \frac{1}{x^{1 / 2}} = x^{- 1 / 2},$ whence $F^{'} (x) = x^{- 1 / 2} = f (x)$ for all $x \in (0, \infty) .$
$F (x) = - 2 \cos x - \frac{3}{5} x^{5 / 3}$ is an antiderivative of $f (x) = 2 \sin x - x^{2 / 3}$ on $I = R$

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Example 1.1.4. Antiderivatives depend on intervals.

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Let

f (x) = {| x |}_{} .

Show that

F (x) = \frac{1}{2} x^{2}

is an antiderivative of

f

I = [0, 1],

but not on

J = [- 1, 0] .

Find an antiderivative of

f

J .

Solution.

We have

F^{'} (x) = x

for all

x \in R .

Since by definition we have

f (x) = {| x |}_{} = {\begin{cases} x & if x \geq 0 \\ - x & x \leq 0 \end{cases},

we see that

F^{'} (x) = f (x) = x

for all

x \in [0, 1] .

Thus

F

is an antiderivative of

f

I .

By the same token, since

F^{'} (x) = x

and

f (x) = - x

J = [- 1, 0],

we see that

F^{'} (x) \neq f (x)

for all

x \in J,

and hence that

F

is not an antiderivative of

f

J .

Lastly, from our analysis above we now see that

G (x) = - \frac{1}{2} x^{2}

is an antiderivative of

f

J = [- 1, 0],

since on this interval we have

G^{'} (x) = f (x) = - x .

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Procedure 1.1.5. Computing antiderivatives by inspection.

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Given a function

f

and interval

I,

to compute (with justification) an antiderivative

F

f

by inspection, proceed as follows.

Provide a candidate antiderivative $F .$
Verify that $F$ is an antiderivative by showing that $F^{'} (x) = f (x)$ for all $x \in I .$

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This is very much a “guess and check” procedure. Do not be shy in hazarding a guess for your candidate antiderivative

F;

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.

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Example 1.1.6. Less elementary antiderivative computations.

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Find an antiderivative for the given function

f

on the given interval

I .

$f (x) = \sec^{2} 5 x,$ $I = (- π / 2, π / 2)$
$f (x) = x \cos (x^{2}),$ $I = R$
$f (x) = \cos (x^{2}),$ $I = R$

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.

$F (x) = \frac{1}{5} \tan 5 x$ is an antiderative of $f (x) = \sec^{2} 5 x$ on $I = (- π / 2, π / 2)$ since

$\begin{aligned} \frac{d}{d x} \frac{1}{5} \tan 5 x & = \frac{1}{5} (\frac{d}{d x} \tan 5 x) & (scalar rule) \\ = \frac{1}{5} \sec^{2} 5 x \cdot (5 x)^{'} & (chain rule) \\ = \frac{1}{5} (5 \sec^{2} 5 x) \\ = \sec^{2} 5 x . \end{aligned}$
$F (x) = \frac{1}{2} \sin (x^{2})$ is an antiderivative of $f (x) = x \cos (x^{2})$ since

$\begin{aligned} F^{'} (x) & = \frac{1}{2} {(\sin (x^{2}))}^{'} & (scalar rule) \\ = \frac{1}{2} \cos (x^{2}) \cdot (x^{2})^{'} & (chain rule) \\ = \frac{1}{2} (2 x \cos (x^{2})) \\ = x \cos (x^{2}) . \end{aligned}$
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!

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Theorem 1.1.7. General antiderivative formula.

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Assume

F

is an antiderivative of

f

on the interval

I .

Given any constant $C \in R,$ the function $G (x) = F (x) + C$ is also an antiderivative of $f$ on $I .$
Conversely, if $G$ is an antiderivative of $f$ on $I,$ then there is a constant $C \in R$ such that $G (x) = F (x) + C$ for all $x \in I .$

Proof.

Let

F

be an antiderivative of

f

on the interval

I

and let

C \in R

be any real constant. We have

\begin{aligned} \frac{d}{d x} (F (x) + C) & = \frac{d}{d x} F (x) + \frac{d}{d x} (C) \\ = F^{'} (x) + 0 \\ = f (x) \end{aligned}

for all

x \in I .

Thus

G (x) = F (x) + C

is an antiderivative for any

C \in R .

Assume

F

and

G

are antiderivatives of

f

on the interval

I,

and define

H (x) = F (x) - G (x) .

We have

H^{'} (x) = F^{'} (x) - G^{'} (x) = f (x) - f (x) = 0

for all

x \in I .

It follows (as a consequence of the mean value theorem) that

H

is a constant function on

I .

In other words, we have

H (x) = C

for all

x \in I .

Since

H (x) = F (x) - G (x),

we conclude that

G (x) = F (x) + H (x) = F (x) + C

for all

x \in I,

as desired.

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Remark 1.1.8.

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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,

then the set of all antiderivatives of

f

I

is given by

{G (x) : G (x) = F (x) + C for some C \in R} .

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In other words, once we find one antiderivative

F

f

I,

we obtain all other antiderivatives of

f

simply by adding an arbitrary constant

C

F .

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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,

then it has infinitely many antiderivatives

F (x) + C,

where

C

is any real constant.

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Example 1.1.9. Leaking water tank.

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At time

t = 0

minutes a 100 gallon tank of water begins leaking. After

t

minutes, the rate at which the gallon leaks is

r (t) = \frac{1}{\sqrt{2 t + 1}} gallons per minute .

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Find a formula for

f (t),

the amount of water in the tank after

t

minutes.

Solution.

Recall that if

Q = f (t),

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 .

We are told that water leaks out of the tank at a rate given by

r (t) .

It follows that

f^{'} (x) = - r (t) = \frac{- 1}{\sqrt{2 t + 1}} .

Put another way,

f

is an antiderivative of

- r (t) = - 1 / \sqrt{2 t + 1} .

An application of Procedure 1.1.5 yields

F (t) = - \sqrt{2 t + 1}

as an antiderivative of

- r (t) .

It follows from Theorem 1.1.7 that the general antiderivative is of the form

F (x) + C = - \sqrt{2 t + 1} + C .

Since

f

is also an antiderivative, we thus have

\begin{matrix} (1.1.1) & f (t) = - \sqrt{2 t + 1} + C . \end{matrix}

To determine what the constant

C

is, we use the initial condition

f (0) = 100 .

Using (1.1.1) this implies that

100 = f (0) = - \sqrt{1} + C = - 1 + C .

We conclude that

C = 101

and thus

f (t) = 101 - \sqrt{2 t + 1} .

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Example 1.1.10. Initial value problem.

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Consider the differential equation

\begin{matrix} (1.1.2) & f^{″} (x) = - \frac{2}{3} \cos 2 x + x . \end{matrix}

Find the general formula for a function $f$ satisfying (1.1.2).
Find the unique function $f$ satisfying (1.1.2) and the initial conditions

$f (0) = 0, f^{'} (0) = - 1 .$

Solution.

First, since $f^{'} (x)$ is an antiderivative of $f^{″} (x) = - \frac{2}{3} \cos 2 x + x,$ using Procedure 1.1.5 and Theorem 1.1.7, we conclude that

$\begin{matrix} (1.1.3) & f^{'} (x) = - \frac{1}{3} \sin 2 x + \frac{1}{2} x^{2} + C \end{matrix}$

for some $C \in R .$ Next, since $f (x)$ is an antiderivative of $f^{'} (x) = - \frac{1}{3} \sin 2 x + \frac{1}{2} x^{2} + C,$ reasoning as above we have

$\begin{matrix} (1.1.4) & f (x) = \frac{1}{6} \cos 2 x + \frac{1}{6} x^{3} + C x + D \end{matrix}$

for some $C, D \in R .$ The general formula for $f^{″}$ is thus given by (1.1.4).
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

$- 1 = f^{'} (0) = 0 + 0 + C,$

which implies $C = - 1 .$ Using this new fact and $f (0) = 0,$ (1.1.4) now implies

$0 = f (0) = \frac{1}{6} + 0 + 0 + D,$

or $D = - 1 / 6 .$ We conclude that

$f (x) = \frac{1}{6} \cos 2 x + \frac{1}{6} x^{3} - x - \frac{1}{6} .$