Introduction to Antiderivatives

Section 4.1 Introduction to Antiderivatives

Subsection 4.1.1 Introduction to Antiderivatives

So far in the course we have learned how to determine the rate of change (i.e. the derivative) of a given function. That is

\begin{matrix} given a function f (x) find \frac{d f}{d x} . \end{matrix}

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Along the way we developed an understanding of limits, which allowed us to define instantaneous rates of change — the derivative. We then went on to develop a number of applications of derivatives to modelling and approximation. In this last section we want to just introduce the idea of antiderivatives. That is

\begin{matrix} given a derivative \frac{d f}{d x} find the original function f (x) . \end{matrix}

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For example — say we know that

\begin{aligned} \frac{d f}{d x} & = x^{2} \end{aligned}

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and we want to find

f (x) .

From our previous experience differentiating we know that derivatives of polynomials are again polynomials. So we guess that our unknown function

f (x)

is a polynomial. Further we know that when we differentiate

x^{n}

we get

n x^{n - 1}

— multiply by the exponent and reduce the exponent by 1. So to end up with a derivative of

x^{2}

we need to have differentiated an

x^{3} .

But

\frac{d}{d x} x^{3} = 3 x^{2},

so we need to divide both sides by 3 to get the answer we want. That is

\begin{aligned} \frac{d}{d x} (\frac{1}{3} x^{3}) & = x^{2} \end{aligned}

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However we know that the derivative of a constant is zero, so we also have

\begin{aligned} \frac{d}{d x} (\frac{1}{3} x^{3} + 1) & = x^{2} \end{aligned}

and

\begin{aligned} \frac{d}{d x} (\frac{1}{3} x^{3} - π) & = x^{2} \end{aligned}

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At this point it will really help the discussion to give a name to what we are doing.

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Definition 4.1.1.

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A function

F (x)

that satisfies

\begin{aligned} \frac{d}{d x} F (x) & = f (x) \end{aligned}

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is called an antiderivative of

f (x) .

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Notice the use of the indefinite article there — an antiderivative. This is precisely because we can always add or subtract a constant to an antiderivative and when we differentiate we’ll get the same answer. We can write this as a lemma, but it is actually just Corollary 2.13.13 (from back in the section on the mean-value theorem) in disguise.

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Lemma 4.1.2.

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Let

F (x)

be an antiderivative of

f (x),

then for any constant

c,

the function

F (x) + c

is also an antiderivative of

f (x) .

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Because of this lemma we typically write antiderivatives with “

+ c

” tacked on the end. That is, if we know that

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

then we would state that the antiderivative of

f (x)

\begin{matrix} F (x) + c \end{matrix}

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where this “

+ c

” is there to remind us that we can always add or subtract some constant and it will still be an antiderivative of

f (x) .

Hence the antiderivative of

x^{2}

\begin{matrix} \frac{1}{3} x^{3} + c \end{matrix}

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Similarly, the antiderivative of

x^{4}

\begin{matrix} \frac{1}{5} x^{5} + c \end{matrix}

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and for

\sqrt{x} = x^{1 / 2}

it is

\begin{matrix} \frac{2}{3} x^{3 / 2} + c \end{matrix}

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This last one is tricky (at first glance) — but we can always check our answer by differentiating.

\begin{aligned} \frac{d}{d x} (\frac{2}{3} x^{3 / 2} + c) & = \frac{2}{3} \cdot \frac{3}{2} x^{1 / 2} + 0 & ✓ \end{aligned}

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Now in order to determine the value of

c

we need more information. For example, we might be asked

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Given that $g^{'} (t) = t^{2}$ and $g (3) = 7$ find $g (t) .$

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We are given the derivative and one piece of additional information and from these two facts we need to find the original function. From our work above we know that

\begin{aligned} g (t) & = \frac{1}{3} t^{3} + c \end{aligned}

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and we can find

c

from the other piece of information

\begin{aligned} 7 = g (3) & = \frac{1}{3} \cdot 27 + c = 9 + c \end{aligned}

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Hence

c = - 2

and so

\begin{aligned} g (t) & = \frac{1}{3} t^{3} - 2 \end{aligned}

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We can then very easily check our answer by recomputing

g (3)

and

g^{'} (t) .

This is a good habit to get into.

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Finding antiderivatives of polynomials is generally not too hard. We just need to use the rule

\begin{matrix} if f (x) = x^{n} then F (x) = \frac{1}{n + 1} x^{n + 1} + c . \end{matrix}

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Of course this breaks down when

n = - 1 .

In order to find an antiderivative for

f (x) = \frac{1}{x}

we need to remember that

\frac{d}{d x} \log x = \frac{1}{x},

and more generally that

\frac{d}{d x} \log | x | = \frac{1}{x} .

See Example 2.10.4. So

\begin{matrix} if f (x) = \frac{1}{x} then F (x) = \log | x | + c \end{matrix}

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Note however, that both

\frac{1}{x}

and

\log | x |

are not defined at

x = 0 .

If you are interested, for example, in functions that are defined all

x \neq 0,

then, for any constants

c_{1}

and

c_{2},

F (x) = {\begin{aligned} \log | x | + c_{1} & x > 0 \\ \log | x | + c_{2} & x < 0 \end{aligned}

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obeys

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

for all

x \neq 0 .

However, typically, in applications, either all of the

x

’s of interest are strictly positive, or all of the

x

’s of interest are negative. In both of those cases, the general

F (x)

\log | x | + c .

So, typically, in tables of antiderivatives, the antiderivative of

1 / x

is listed as

\log | x | + c .

Example 4.1.3. Antiderivative of $3 x^{5} - 7 x^{2} + 2 x + 3 + x^{- 1} - x^{- 2}$ .

Let

f (x) = 3 x^{5} - 7 x^{2} + 2 x + 3 + x^{- 1} - x^{- 2} .

Then the antiderivative of

f (x)

\begin{aligned} F (x) & = \frac{3}{6} x^{6} - \frac{7}{3} x^{3} + \frac{2}{2} x^{2} + 3 x + \log | x | - \frac{1}{- 1} x^{- 1} + c & clean it up \\ = \frac{1}{2} x^{6} - \frac{7}{3} x^{3} + x^{2} + 3 x + \log | x | + x^{- 1} + c \end{aligned}

Now to check we should differentiate and hopefully we get back to where we started

\begin{aligned} F^{'} (x) & = \frac{6}{2} x^{5} - \frac{7}{3} \cdot 3 x^{2} + 2 x + 3 + \frac{1}{x} - x^{- 2} \\ = 3 x^{5} - 7 x^{2} + 2 x + 3 + \frac{1}{x} - x^{- 2} & ✓ \end{aligned}

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In your next calculus course you will develop a lot of machinery to help you find antiderivatives. At this stage about all that we can do is continue the sort of thing we have done. Think about the derivatives we know and work backwards. So, for example, we can take a list of derivatives

$F (x)$	$1$	$x^{n}$	$\sin x$	$\cos x$	$\tan x$	$e^{x}$	$\ln \| x \|$	$\arcsin x$	$\arctan x$
$f (x) = \frac{d}{d x} F (x)$	$0$	$n x^{n - 1}$	$\cos x$	$- \sin x$	$\sec^{2} x$	$e^{x}$	$\frac{1}{x}$	$\frac{1}{\sqrt{1 - x^{2}}}$	$\frac{1}{1 + x^{2}}$

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and flip it upside down to give the tables of antiderivatives.

$f (x) = \frac{d}{d x} F (x)$	$0$	$n x^{n - 1}$	$\cos x$	$- \sin x$	$\sec^{2} x$	$e^{x}$	$\frac{1}{x}$
$F (x)$	$c$	$x^{n} + c$	$\sin x + c$	$\cos x + c$	$\tan x + c$	$e^{x} + c$	$\ln \| x \| + c$

$f (x) = \frac{d}{d x} F (x)$	$\frac{1}{\sqrt{1 - x^{2}}}$	$\frac{1}{1 + x^{2}}$
$F (x)$	$\arcsin x + c$	$\arctan x + c$

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Here

c

is just a constant — any constant. But we can do a little more; clean up

x^{n}

by dividing by

n

and then replacing

n

n + 1 .

Similarly we can tweak

\sin x

by multiplying by

- 1 :

$f (x) = \frac{d}{d x} F (x)$	$0$	$x^{n}$	$\cos x$	$\sin x$	$\sec^{2} x$	$e^{x}$	$\frac{1}{x}$
$F (x)$	$c$	$\frac{1}{n + 1} x^{n + 1} + c$	$\sin x + c$	$- \cos x + c$	$\tan x + c$	$e^{x} + c$	$\ln \| x \| + c$

$f (x) = \frac{d}{d x} F (x)$	$\frac{1}{\sqrt{1 - x^{2}}}$	$\frac{1}{1 + x^{2}}$
$F (x)$	$\arcsin x + c$	$\arctan x + c$

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Here are a couple more examples.

Example 4.1.4. Antiderivatives of $\sin x, \cos 2 x$ and $\frac{1}{1 + 4 x^{2}}$ .

Consider the functions

\begin{aligned} f (x) & = \sin x + \cos 2 x & g (x) & = \frac{1}{1 + 4 x^{2}} \end{aligned}

Find their antiderivatives.

Solution The first one we can almost just look up our table. Let

F

be the antiderivative of

f,

then

\begin{aligned} F (x) & = - \cos x + \sin 2 x + c & is not quite right . \end{aligned}

When we differentiate to check things, we get a factor of two coming from the chain rule. Hence to compensate for that we multiply

\sin 2 x

\frac{1}{2} :

\begin{aligned} F (x) & = - \cos x + \frac{1}{2} \sin 2 x + c \end{aligned}

Differentiating this shows that we have the right answer.

Similarly, if we use

G

to denote the antiderivative of

g,

then it appears that

G

is nearly

\arctan x .

To get this extra factor of

4

we need to substitute

x \mapsto 2 x .

So we try

\begin{aligned} G (x) & = \arctan (2 x) + c & which is nearly correct . \end{aligned}

Differentiating this gives us

\begin{aligned} G^{'} (x) & = \frac{2}{1 + (2 x)^{2}} = 2 g (x) \end{aligned}

Hence we should multiply by

\frac{1}{2} .

This gives us

\begin{aligned} G (x) & = \frac{1}{2} \arctan (2 x) + c . \end{aligned}

We can then check that this is, in fact, correct just by differentiating.

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Now let’s do a more substantial example.

Example 4.1.5. Position as antiderivative of velocity.

Suppose that we are driving to class. We start at

x = 0

at time

t = 0 .

Our velocity is

v (t) = 50 \sin (t) .

The class is at

x = 25 .

When do we get there?

Solution Let’s denote by

x (t)

our position at time

t .

We are told that

$x (0) = 0$
$x^{'} (t) = 50 \sin t$

We have to determine

x (t)

and then find the time

T

that obeys

x (T) = 25 .

Now armed with our table above we know that the antiderivative of

\sin t

is just

- \cos t .

We can check this:

\begin{aligned} \frac{d}{d t} (- \cos t) & = \sin t \end{aligned}

We can then get the factor of

50

by multiplying both sides of the above equation by 50:

\begin{aligned} \frac{d}{d t} (- 50 \cos t) & = 50 \sin t \end{aligned}

And of course, this is just an antiderivative of

50 \sin t;

to write down the general antiderivative we just add a constant

c :

\begin{aligned} \frac{d}{d t} (- 50 \cos t + c) & = 50 \sin t \end{aligned}

Since

v (t) = \frac{d}{d t} x (t),

this antiderivative is

x (t) :

\begin{aligned} x (t) & = - 50 \cos t + c \end{aligned}

To determine

c

we make use of the other piece of information we are given, namely

\begin{aligned} x (0) & = 0. \end{aligned}

Substituting this in gives us

\begin{aligned} x (0) & = - 50 \cos 0 + c = - 50 + c \end{aligned}

Hence we must have

c = 50

and so

\begin{aligned} x (t) & = - 50 \cos t + 50 = 50 (1 - \cos t) . \end{aligned}

Now that we have our position as a function of time, we can determine how long it takes us to arrive there. That is, we can find the time

T

so that

x (T) = 25 .

\begin{aligned} 25 = x (T) & = 50 (1 - \cos T) & so \\ \frac{1}{2} & = 1 - \cos T \\ - \frac{1}{2} & = - \cos T \\ \frac{1}{2} & = \cos T . \end{aligned}

Recalling our special triangles, we see that

T = \frac{π}{3} .

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The example below shows how antiderivatives arise naturally when studying differential equations.

Example 4.1.6. Theorem 3.3.2 revisited..

Back in Section 3.3 we encountered a simple differential equation, namely equation 3.3.1. We were able to solve this equation by guessing the answer and then checking it carefully. We can derive the solution more systematically by using antiderivatives.

Recall equation 3.3.1:

\begin{aligned} \frac{d Q}{d t} & = - k Q \end{aligned}

where

Q (t)

is the amount of radioactive material at time

t

and we assume

Q (t) > 0 .

Take this equation and divide both sides by

Q (t)

to get

\begin{aligned} \frac{1}{Q (t)} \frac{d Q}{d t} & = - k \end{aligned}

At this point we should

Well — perhaps it is better to say “notice that”. Let’s not make this a moral point.

think that the left-hand side is familiar. Now is a good moment to look back at logarithmic differentiation in Section 2.10.

The left-hand side is just the derivative of

\log Q (t) :

\begin{aligned} \frac{d}{d t} (\log Q (t)) & = \frac{1}{Q (t)} \frac{d Q}{d t} \\ = - k \end{aligned}

So to solve this equation, we are really being asked to find all functions

\log Q (t)

having derivative

- k .

That is, we need to find all antiderivatives of

- k .

Of course that is just

- k t + c .

Hence we must have

\begin{aligned} \log Q (t) & = - k t + c \end{aligned}

and then taking the exponential of both sides gives

\begin{aligned} Q (t) & = e^{- k t + c} = e^{c} \cdot e^{- k t} = C e^{- k t} \end{aligned}

where

C = e^{c} .

This is precisely Theorem 3.3.2.

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The above is a small example of the interplay between antiderivatives and differential equations.

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Here is another example of how we might use antidifferentiation to compute areas or volumes.

Example 4.1.7. Volume of a cone.

We know (especially if one has revised the material in the appendix and Appendix B.5.2 in particular) that the volume of a right-circular cone is

\begin{aligned} V & = \frac{π}{3} r^{2} h \end{aligned}

where

h

is the height of the cone and

r

is the radius of its base. Now, the derivation of this formula given in Appendix B.5.2 is not too simple. We present an alternate proof here that uses antiderivatives.

Consider cutting off a portion of the cone so that its new height is

x

(rather than

h

). Call the volume of the resulting smaller cone

V (x) .

We are going to determine

V (x)

for all

x \geq 0,

including

x = h,

by first evaluating

V^{'} (x)

and

V (0)

(which is obviously

0

Call the radius of the base of the new smaller cone

y

(rather than

r

). By similar triangles we know that

\begin{aligned} \frac{r}{h} & = \frac{y}{x} . \end{aligned}

Now keep

x

and

y

fixed and consider cutting off a little more of the cone so its height is

X .

When we do so, the radius of the base changes from

y

Y

and again by similar triangles we know that

\begin{aligned} \frac{Y}{X} & = \frac{y}{x} = \frac{r}{h} \end{aligned}

The change in volume is then

\begin{matrix} V (x) - V (X) \end{matrix}

Of course if we knew the formula for the volume of a cone, then we could compute the above exactly. However, even without knowing the volume of a cone, it is easy to derive upper and lower bounds on this quantity. The piece removed has bottom radius

y

and top radius

Y .

Hence its volume is bounded above and below by the cylinders of height

x - X

and with radius

y

and

Y

respectively. Hence

\begin{aligned} π Y^{2} (x - X) & \leq V (x) - V (X) \leq π y^{2} (x - X) \end{aligned}

since the volume of a cylinder is just the area of its base times its height. Now massage this expression a little

\begin{aligned} π Y^{2} & \leq \frac{V (x) - V (X)}{x - X} \leq π y^{2} \end{aligned}

The middle term now looks like a derivative; all we need to do is take the limit as

X \to x :

\begin{aligned} lim_{X \to x} π Y^{2} & \leq lim_{X \to x} \frac{V (x) - V (X)}{x - X} \leq lim_{X \to x} π y^{2} \end{aligned}

The rightmost term is independent of

X

and so is just

π y^{2} .

In the leftmost term, as

X \to x,

we must have that

Y \to y .

Hence the leftmost term is just

π y^{2} .

Then by the squeeze theorem (Theorem 1.4.18) we know that

\begin{aligned} \frac{d V}{d x} = lim_{X \to x} \frac{V (x) - V (X)}{x - X} & = π y^{2} . \end{aligned}

But we know that

\begin{aligned} y & = \frac{r}{h} \cdot x \end{aligned}

\begin{aligned} \frac{d V}{d x} & = π {(\frac{r}{h})}^{2} x^{2} \end{aligned}

Now we can antidifferentiate to get back to

V :

\begin{aligned} V (x) & = \frac{π}{3} {(\frac{r}{h})}^{2} x^{3} + c \end{aligned}

To determine

c

notice that when

x = 0

the volume of the cone is just zero and so

c = 0 .

Thus

\begin{aligned} V (x) & = \frac{π}{3} {(\frac{r}{h})}^{2} x^{3} \end{aligned}

and so when

x = h

we are left with

\begin{aligned} V (h) & = \frac{π}{3} {(\frac{r}{h})}^{2} h^{3} = \frac{π}{3} r^{2} h \end{aligned}

as required.

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Let

f (x)

be a function with derivative

f^{'} (x) .

What is the most general antiderivative of

f^{'} (x) ?

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2.

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On the graph below, the black curve is

y = f (x) .

Which of the coloured curves is an antiderivative of

f (x) ?

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Exercises — Stage 2 .

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In Questions 4.1.2.3 through 4.1.2.12, you are asked to find the antiderivative of a function. Phrased like this, we mean the most general antiderivative. These will all include some added constant. The table after Example 4.1.3 might be of help.

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Find the antiderivative of

f (x) = \frac{1}{1 + 25 x^{2}}

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13.

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Find the function

f (x)

with

f^{'} (x) = 3 x^{2} - 9 x + 4

and

f (1) = 10 .

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14.

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Find the function

f (x)

with

f^{'} (x) = \cos (2 x)

and

f (π) = π .

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15.

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Find the function

f (x)

that is defined and obeys

f^{'} (x) = \frac{1}{x}

for all

x < 0,

and also obeys

f (- 1) = 0 .

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16.

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Find the function

f (x)

with

f^{'} (x) = \frac{1}{\sqrt{1 - x^{2}}} + 1

and

f (1) = - \frac{π}{2} .

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17.

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Suppose a population of bacteria at time

t

(measured in hours) is growing at a rate of

100 e^{2 t}

individuals per hour. Starting at time

t = 0,

how long will it take the initial colony to increase by 300 individuals?

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18.

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Your bank account at time

t

(measured in years) is growing at a rate of

1500 e^{\frac{t}{50}}

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dollars per year. How much money is in your account at time

t ?

🔗

19.

🔗

At time

t

during a particular day,

0 \leq t \leq 24,

your house consumes energy at a rate of

0.5 \sin (\frac{π}{24} t) + 0.25

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kW. (Your consumption was smallest in the middle of the night, and peaked at noon.) How much energy did your house consume in that day?

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Exercises — Stage 3 .

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For Questions 4.1.2.21 through 4.1.2.26, you are again asked to find the antiderivatives of certain functions. In general, finding antiderivatives can be extremely difficult--indeed, it will form the main topic of next semester’s calculus course. However, you can work out the antiderivatives of the functions below using what you’ve learned so far about derivatives.

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20. (✳).

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Let

f (x) = 2 \sin^{- 1} \sqrt{x}

and

g (x) = \sin^{- 1} (2 x - 1) .

Find the derivative of

f (x) - g (x)

and simplify your answer. What does the answer imply about the relation between

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Find the antiderivative of

f (x) = \frac{7}{\sqrt{3 - x^{2}}} .

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27.

🔗

Imagine forming a solid by revolving the parabola

y = x^{2} + 1

around the

x

-axis, as in the picture below.

🔗

Use the method of Example 4.1.7 to find the volume of such an object if the segment of the parabola that we rotate runs from

x = - H

x = H .

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28.

🔗

Find all functions

f (x)

that are defined and obey

f^{'} (x) = \frac{1}{x}

for all

x \neq 0,

and also obey

f (- 1) = 0 .

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Section 4.1 Introduction to Antiderivatives

Subsection 4.1.1 Introduction to Antiderivatives

Definition 4.1.1.

Lemma 4.1.2.

Example 4.1.3. Antiderivative of 3x5−7x2+2x+3+x−1−x−2.

Example 4.1.4. Antiderivatives of sin⁡x,cos⁡2x and 11+4x2.

Example 4.1.5. Position as antiderivative of velocity.

Example 4.1.6. Theorem 3.3.2 revisited..

Example 4.1.7. Volume of a cone.

Exercises 4.1.2 Exercises

Exercises — Stage 1 .

1.

2.

Exercises — Stage 2 .

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

Exercises — Stage 3 .

20. (✳).

21.

22.

23.

24.

25.

26.

27.

28.

Example 4.1.3. Antiderivative of $3 x^{5} - 7 x^{2} + 2 x + 3 + x^{- 1} - x^{- 2}$ .

Example 4.1.4. Antiderivatives of $\sin x, \cos 2 x$ and $\frac{1}{1 + 4 x^{2}}$ .