3.9 Antiderivatives
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Antiderivatives
A physicist who knows the velocity of a particle might wish to know its position at a given time.
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.
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.
Antiderivatives
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.
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Antiderivatives
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) = x3, then F′(x) = x2 = f(x).
But the function G(x) = x3 + 100 also satisfies G′(x) = x2. Therefore both F and G are antiderivatives of f.
Indeed, any function of the form H(x) = x3 + C, where C is a constant, is an antiderivative of f.
Antiderivatives
The following theorem says that f has no other antiderivative.
Going back to the function f(x) = x2, we see that the general antiderivative of f is x3 + C.
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Antiderivatives
By assigning specific values to the constant C, we obtain a family of functions whose graphs are vertical translates of one another (see Figure 1).
This makes sense because each curve must have the same slope at any given value of x.
Figure 1
Members of the family of antiderivatives of f(x) = x2
Example 1
Find the most general antiderivative of each of the following functions.
(a) f(x) = sin x (b) f(x) = xn, n ≥ 0 (c) f(x) = x–3 Solution:
(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.
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Example 1 – Solution
(b) We use the Power Rule to discover an antiderivative of xn:
Therefore the general antiderivative of f(x) = xn is
This is valid for n ≥ 0 because then f(x) = xn is defined on an interval.
cont’d
Example 1 – Solution
(c) If we put n = – 3 in the antiderivative formula from part (b), we get the particular antiderivative
F(x) = x–2/(– 2).
But notice that f(x) = x–3 is not defined at x = 0.
Thus Theorem 1 tells us only that the general
antiderivative of f is x–2/(– 2) + C on any interval that does not contain 0. So the general antiderivative of f(x) = 1/x3 is
cont’d
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Antiderivatives
As in Example 1, every differentiation formula, when read from right to left, gives rise to an antidifferentiation formula.
In Table 2 we list some particular antiderivatives.
To obtain the most general antiderivative from the particular ones in Table 2, we have to add a constant (or constants), as in Example 1.
Antiderivatives
Each formula in the table is true because the derivative of the function in the right column appears in the left column.
In particular, the first formula says that the antiderivative of a constant times a function is the constant times the
antiderivative of the function.
The second formula says that the antiderivative of a sum is
the sum of the antiderivatives. (We use the notation F′ = f, G′ = g.)
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Antiderivatives
An equation that involves the derivatives of a function is called a differential equation.
The general solution of a differential equation involves an arbitrary constant (or constants).
However, there may be some extra conditions given that will determine the constants and therefore uniquely specify the solution.
Rectilinear Motion
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Rectilinear Motion
Antidifferentiation is particularly useful in analyzing the
motion of an object moving in a straight line. We know 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.
Likewise, the acceleration function is a(t) = v′(t), so the velocity function is an antiderivative of the acceleration.
If the acceleration and the initial values s(0) and v(0) are known, then the position function can be found by
antidifferentiating twice.
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).
Solution:
Since v′(t) = a(t) = 6t + 4, antidifferentiation gives v(t) = 6 + 4t + C
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Example 6 – Solution
Note that v(0) = C. But we are given that v(0) = –6, so C = –6 and
v(t) = 3t2 + 4t – 6
Since v(t) = s′(t), s is the antiderivative of v:
s(t) = 3 + 4 – 6t + D
This gives s(0) = D. We are given that s(0) = 9, so D = 9 and the required position function is
s(t) = t3 + 2t2 – 6t + 9
= t3 + 2t2 – 6t + D
cont’d
