Contributed by:
In this section, we will learn about: Antiderivatives and how they are useful in solving certain scientific problems.
1.
4
APPLICATIONS OF DIFFERENTIATION
2.
APPLICATIONS OF DIFFERENTIATION
4.9
Antiderivatives
In this section, we will learn about:
Antiderivatives and how they are useful
in solving certain scientific problems.
3.
A physicist who knows the velocity of
a particle might wish to know its position
at a given time.
4.
An engineer who can measure the variable
rate at which water is leaking from a tank
wants to know the amount leaked over
a certain time period.
5.
A biologist who knows the rate at which
a bacteria population is increasing might want
to deduce what the size of the population will
be at some future time.
6.
In each case, the problem is to find
a function F whose derivative is a known
function f.
If such a function F exists, it is called
an antiderivative of f.
7.
A function F is called an antiderivative
of f on an interval I if F’(x) = f(x) for all
x in I.
8.
For instance, let f(x) = x2.
It isn’t difficult to discover an antiderivative of f
if we keep the Power Rule in mind.
In fact, if F(x) = ⅓ x3, then F’(x) = x2 = f(x).
9.
However, the function G(x) = ⅓ x3 + 100
also satisfies G’(x) = x2.
Therefore, both F and G are antiderivatives of f.
10.
Indeed, any function of the form
H(x)=⅓ x3 + C, where C is a constant,
is an antiderivative of f.
The question arises: Are there any others?
11.
To answer the question, recall that,
in Section 4.2, we used the Mean Value
It was to prove that, if two functions have identical
derivatives on an interval, then they must differ by
a constant (Corollary 7).
12.
Thus, if F and G are any two antiderivatives
of f, then F’(x) = f(x) = G(x)
So, G(x) – F(x) = C, where C is a constant.
We can write this as G(x) = F(x) + C.
Hence, we have the following theorem.
13.
ANTIDERIVATIVES Theorem 1
If F is an antiderivative of f on an interval I,
the most general antiderivative of f on I is
F(x) + C
where C is an arbitrary constant.
14.
Going back to the function f(x) = x2,
we see that the general antiderivative
of f is ⅓ x3 + C.
15.
FAMILY OF FUNCTIONS
By assigning specific values to C, we
obtain a family of functions.
Their graphs are
vertical translates
of one another.
This makes sense, as
each curve must have
the same slope at any
given value of x.
16.
ANTIDERIVATIVES Example 1
Find the most general antiderivative of
each function.
a. f(x) = sin x
b. f(x) = 1/x
c. f(x) = xn, n ≠ -1
17.
ANTIDERIVATIVES Example 1 a
If f(x) = - cos x, then F’(x) = sin x.
So, an antiderivative of sin x is - cos x.
By Theorem 1, the most general antiderivative
is: G(x) = - cos x + C
18.
ANTIDERIVATIVES Example 1 b
Recall from Section 3.6 that
d 1
(ln x)
dx x
So, on the interval (0, ∞), the general antiderivative
of 1/x is ln x + C.
19.
ANTIDERIVATIVES Example 1 b
We also learned that
d 1
(ln | x |)
dx x
for all x ≠ 0.
Theorem 1 then tells us that the general antiderivative
of f(x) = 1/x is ln |x| + C on any interval that doesn’t
contain 0.
20.
ANTIDERIVATIVES Example 1 b
In particular, this is true on each of
the intervals (-∞, 0) and (0, ∞).
So, the general antiderivative of f is:
ln x C1 if x 0
F ( x)
ln( x) C2 if x 0
21.
ANTIDERIVATIVES Example 1 c
We use the Power Rule to discover
an antiderivative of xn.
In fact, if n ≠ -1, then
n 1 n
d x (n 1) x n
x
dx n 1 n 1
22.
ANTIDERIVATIVES Example 1 c
Therefore, the general antiderivative
n 1
of f(x) = xn is: x
F ( x) C
n 1
This is valid for n ≥ 0 since then f(x) = xn is defined
on an interval.
If n is negative (but n ≠ -1), it is valid on any interval
that doesn’t contain 0.
23.
ANTIDERIVATIVE FORMULA
As in the example, every differentiation
formula, when read from right to left,
gives rise to an antidifferentiation formula.
24.
ANTIDERIVATIVE FORMULA Table 2
Here, we list some particular
25.
ANTIDERIVATIVE FORMULA
Each formula is true because the derivative
of the function in the right column appears
in the left column.
26.
ANTIDERIVATIVE FORMULA
In particular, the first formula says that the
antiderivative of a constant times a function
is the constant times the antiderivative of
the function.
27.
ANTIDERIVATIVE FORMULA
The second formula says that the
antiderivative of a sum is the sum of the
We use the notation F’ = f, G’ = g.
28.
ANTIDERIVATIVES Example 2
Find all functions g such that
5
2x x
g '( x) 4sin x
x
29.
ANTIDERIVATIVES Example 2
First, we rewrite the given function:
5
2x x 4 1
g '( x) 4sin x 4sin x 2 x
x x x
Thus, we want to find an antiderivative of:
4 1 2
g '( x) 4sin x 2 x x
30.
ANTIDERIVATIVES Example 2
Using the formulas in Table 2 together
with Theorem 1, we obtain:
5 12
x x
g ( x) 4( cos x) 2 1 C
5 2
5
4 cos x x 2 x C
2
5
31.
In applications of calculus, it is very common
to have a situation as in the example—
where it is required to find a function, given
knowledge about its derivatives.
32.
DIFFERENTIAL EQUATIONS
An equation that involves the derivatives of
a function is called a differential equation.
These will be studied in some detail in Chapter 9.
For the present, we can solve some elementary
differential equations.
33.
DIFFERENTIAL EQUATIONS
The general solution of a differential equation
involves an arbitrary constant (or constants),
as in Example 2.
However, there may be some extra conditions
given that will determine the constants and, therefore,
uniquely specify the solution.
34.
DIFFERENTIAL EQUATIONS Example 3
Find f if
f’(x) = ex + 20(1 + x2)-1 and f(0) = – 2
The general antiderivative of
x 20
f '( x) e
1 x2
x 1
is f '( x) e 20 tan x C
35.
DIFFERENTIAL EQUATIONS Example 3
To determine C, we use the fact that
f(0) = – 2 :
f(0) = e0 + 20 tan-10 + C = – 2
Thus, we have: C = – 2 – 1 = – 3
So, the particular solution is:
f(x) = ex + 20 tan-1x –
3
36.
DIFFERENTIAL EQUATIONS Example 4
Find f if f’’(x) = 12x2 + 6x – 4,
f(0) = 4, and f(1) = 1.
37.
DIFFERENTIAL EQUATIONS Example 4
The general antiderivative of
f’’(x) = 12x2 + 6x – 4 is:
3 2
x x
f '( x) 12 6 4 x C
3 2
3 2
4 x 3x 4 x C
38.
DIFFERENTIAL EQUATIONS Example 4
Using the antidifferentiation rules once
more, we find that:
4 3 2
x x x
f ( x) 4 3 4 Cx D
4 3 2
4 3 2
x x 2 x Cx D
39.
DIFFERENTIAL EQUATIONS Example 4
To determine C and D, we use the given
conditions that f(0) = 4 and f(1) = 1.
As f(0) = 0 + D = 4, we have: D = 4
As f (1) = 1 + 1 – 2 + C + 4 = 1, we have: C = –3
40.
DIFFERENTIAL EQUATIONS Example 4
Therefore, the required function
f(x) = x4 + x3 – 2x2 – 3x + 4
41.
If we are given the graph of a function f,
it seems reasonable that we should be able
to sketch the graph of an antiderivative F.
42.
Suppose we are given that F(0) = 1.
We have a place to start—the point (0, 1).
The direction in which we move our pencil
is given at each stage by the derivative F’(x) = f(x).
43.
In the next example, we use the principles
of this chapter to show how to graph F even
when we don’t have a formula for f.
This would be the case, for instance, when
f(x) is determined by experimental data.
44.
GRAPH Example 5
The graph of a function f is given.
Make a rough sketch of an antiderivative F,
given that F(0) = 2.
45.
GRAPH Example 5
We are guided by the fact that
the slope of y = F(x) is f(x).
46.
GRAPH Example 5
We start at (0, 2) and
draw F as an initially
decreasing function
since f(x) is negative
when 0 < x < 1.
47.
GRAPH Example 5
Notice f(1) = f(3) = 0.
So, F has horizontal
tangents when x = 1
and x = 3.
For 1 < x < 3, f(x)
is positive.
Thus, F is increasing.
48.
GRAPH Example 5
We see F has a local
minimum when x = 1
and a local maximum
when x = 3.
For x > 3, f(x) is
negative.
Thus, F is decreasing
on (3, ∞).
49.
GRAPH Example 5
Since f(x) → 0 as
x → ∞, the graph
of F becomes flatter
as x → ∞.
50.
GRAPH Example 5
Also, F’’(x) = f’(x)
changes from positive
to negative at x = 2
and from negative to
positive at x = 4.
So, F has inflection
points when x = 2
and x = 4.
51.
RECTILINEAR MOTION
Antidifferentiation is particularly useful
in analyzing the motion of an object
moving in a straight line.
52.
RECTILINEAR MOTION
Recall that, if the object has position
function s = f(t), then the velocity function
is v(t) = s’(t).
This means that the position function is
an antiderivative of the velocity function.
53.
RECTILINEAR MOTION
Likewise, the acceleration function
is a(t) = v’(t).
So, the velocity function is an antiderivative
of the acceleration.
54.
RECTILINEAR MOTION
If the acceleration and the initial values
s(0) and v(0) are known, then the position
function can be found by antidifferentiating
55.
RECTILINEAR MOTION Example 6
A particle moves in a straight line and has
acceleration given by a(t) = 6t + 4.
Its initial velocity is v(0) = -6 cm/s and its initial
displacement is s(0) = 9 cm.
Find its position function s(t).
56.
RECTILINEAR MOTION Example 6
As v’(t) = a(t) = 6t + 4, antidifferentiation
2
t
v(t ) 6 4t C
2
2
3t 4t C
57.
RECTILINEAR MOTION Example 6
Note that v(0) = C.
However, we are given that v(0) = – 6,
so C = – 6.
Therefore, we have:
v(t) = 3t2 + 4t – 6
58.
RECTILINEAR MOTION Example 6
As v(t) = s’(t), s is the antiderivative
of v: t 3
t 2
s (t ) 3 4 6t D
3 2
3 2
t 2t 6t D
This gives s(0) = D.
We are given that s(0) = 9, so D = 9.
59.
RECTILINEAR MOTION Example 6
The required position function
s(t) = t3 + 2t 2 – 6t + 9
60.
RECTILINEAR MOTION
An object near the surface of the earth is
subject to a gravitational force that produces
a downward acceleration denoted by g.
For motion close to the ground, we may
assume that g is constant.
Its value is about 9.8 m/s2 (or 32 ft/s2).
61.
RECTILINEAR MOTION Example 7
A ball is thrown upward with a speed of
48 ft/s from the edge of a cliff 432 ft above
the ground.
Find its height above the ground t seconds later.
When does it reach its maximum height?
When does it hit the ground?
62.
RECTILINEAR MOTION Example 7
The motion is vertical, and we choose
the positive direction to be upward.
At time t, the distance above the ground is s(t)
and the velocity v(t) is decreasing.
So, the acceleration must be negative
and we have: dv
a (t ) 32
dt
63.
RECTILINEAR MOTION Example 7
Taking antiderivatives, we have
v(t) = – 32t + C
To determine C, we use the information
that v(0) = 48.
This gives: 48 = 0 + C
So, v(t) = – 32t + 48
64.
RECTILINEAR MOTION Example 7
The maximum height is reached
v(t) = 0, that is, after 1.5 s
65.
RECTILINEAR MOTION Example 7
As s’(t) = v(t), we antidifferentiate again
and obtain:
s(t) = – 16t2 + 48t + D
Using the fact that s(0) = 432, we have
432 = 0 + D. So,
s(t) = – 16t2 + 48t + 432
66.
RECTILINEAR MOTION Example 7
The expression for s(t) is valid until the ball
hits the ground.
This happens when s(t) = 0, that is, when
– 16t2 + 48t + 432 = 0
Equivalently: t2 – 3t – 27 = 0
67.
RECTILINEAR MOTION Example 7
Using the quadratic formula to solve this
equation, we get: 3 3 13
t
2
We reject the solution with the minus sign—as
it gives a negative value for t.
68.
RECTILINEAR MOTION Example 7
Therefore, the ball hits the ground
3(1 + 13 )/2 ≈ 6.9 s
+1-315-636-4485
+91 98151-74050 
