Antiderivatives
Suppose that you know the velocity of an object, v (t). How do you find
the position at a given time? Find a function x(t) such that
x 0 (t) = v (t).
Suppose that you know the outflow from a reservoir, through a dam r (t).
How do you know how much water has been released over a certain
period, from t1 to t2 ? Find a function w (t) such that w 0 (t) = r (t) and
consider the quantity w (t2 ) − w (t1 ).
Suppose that you know the rate at which a population of bacteria
increases r (t). What will the size of the population be in the future?
Find a function p(t) such that p 0 (t) = r (t).
Suppose that you know the density ρ(t) of a linear object, 0 ≤ t ≤ 2.
What is the total mass of the object? Find a function F (t) such that
F 0 (t) = ρ(t), then the total mass is given by F 0 (2) − F 0 (0).
In each case, we’re trying to “undo” the differentiation process.
Antiderivatives
antiderivative
A function F is called an antiderivative of f on an interval I if
F 0 (x) = f (x) for all x in I .
Example: Find an antiderivative for f (x) = x 2 . Want to try an undo the
power rule.
d 3
[x ] = 3x 2
dx
So x 3 almost works except for the coefficient 3. We can cancel that by
mulitplying by 1/3. Try F (x) = x 3 /3:
F 0 (x) = (1/3)3x 2 = x 2 = f (x)
Antiderivatives
But adding a constant does not change the derivative:
d 3
[x /3 + π] = x 2
dx
d 3
[x /3 + 23] = x 2
dx
d 3
[x /3 − e] = x 2
dx
d 3
[x /3 + 42] = x 2
dx
Theorem
If F is an antiderivative of f on an interval I , then the most general
antiderivative of f on an interval I is F (x) + C , where C is an arbitrary
constant.
Exercise
Find the most general antiderivative of g (x) =
1 + t + t2
√
t
2
2
SOLUTION: F (x) = 2t 1/2 + t 3/2 + t 5/2 + C
3
5
Antiderivatives
Suppose that we add one more piece of information: the value of the
antiderivative F at a point. Then the antiderivative is specified
completely.
Example: Find the antiderivative F of f (x) = x + 2 sin x that satisfies
F (0) = −6
Solution: The most general antiderivative is
F (x) =
1 2
x − 2 cos x + C
2
Then F (0) = −2 + C , so we must have C = −4. Thus antiderivative is
F (x) =
1 2
x − 2 cos x − 4
2
Exercise 1
Exercise 1: A particle is moving with velocity given by
v (t) = sin t − cos t. What is the position of the particle as a function of
time, s(t), if its position at time zero is s(0) = 0?
Solution: First find the most general antiderivative for v (t) and call it
s(t).
s(t) = − cos t − sin t + C
Now use the fact that the particle starts at position zero:
0 = s(0) = −1 + C
So C = 1 and the answer the position of the particle is given by:
s(t) = − cos t − sin t + 1
Exercise 2
Exercise 2: A particle is moving with acceleration given by
a(t) = t 2 − 4t + 6. What is the position of the particle as a function of
time, s(t), if its position at time zero is s(0) = 0 and its position at time
1 is s(1) = 20?
Solution: The antiderivative of acceleration is velocity, so
v (t) =
1 3
t − 2t 2 + 6t + C
3
where C is some constant that is yet to be determined. The
antiderivative of velocity is position, so
s(t) =
1 4 2 3
t − t + 3t 2 + Ct + D
12
3
where D is some constant yet to be determined.
Exercise 2 (cont.)
Using the last piece of information from the problem, the positions at
time 0 and time 1, we have
0 = s(0) = C
20 = s(1) =
2
5
1
− +3+D =
+D
12 3
12
or D = 235/12. Thus the position function is
s(t) =
1 4 2 3
235
t − t + 3t 2 +
12
3
12