Antiderivatives
Definition A function, F, is said to be an antiderivative of a function, f, on an interval, I, if
F  x  fx for all x  I.
Theorem If F is an antiderivative of f on I, then every function of the form F  C (where C
can be any constant) is also an antiderivative of f on I. Furthermore, the set of functions of
the form F  C (where C can be any constant) comprise all of the antiderivatives of f on I.
Example Find all antiderivatives of the following functions, f, on the given intervals, I.
1.
2.
3.
4.
5.
6.
7.
8.
fx
fx
fx
fx
fx
fx
fx
fx








cosx on the interval , .
secx tanx on the interval  2 , 2 .
x 5 on the interval , .
x 1/2 on the interval 0, 
e x on the interval , 
4x 2  6x  3 on the interval , 
x 1 on the interval 0, .
x 1 on the interval , 0
Solution
1. We know that
d sinx  cosx for all x  , .
dx
Therefore, the function Fx  sinx is an antiderivative of the function fx  cosx on
the interval , . This means that every antiderivative of f on the interval ,  has
the form
Gx  sinx  C (where C can be any constant).
Remark If a function, f, has an antiderivative on some interval, I, then f actually has
infinitely many antiderivatives on I. However, if we specify that we are looking for a
particular antiderivative of f that has a certain given value at a certain point in I, then this
narrows the choice down to just one.
Example Find the particular antiderivative, F, of each the following functions, f, on the
given intervals, I, that satisfies the given requirement.
1.
2.
3.
4.
5.
6.
7.
8.
fx
fx
fx
fx
fx
fx
fx
fx








cosx on the interval , , F0  6.
secx tanx on the interval  2 , 2 , F0  0.
x 5 on the interval , , F2  6.
x 1/2 on the interval 0, , F1  1.
e x on the interval , , F0  3.
4x 2  6x  3 on the interval , , F3  0.
x 1 on the interval 0, , F1  4.
x 1 on the interval , 0, F2  7.
Solution
1. Every antiderivative, F, of the function fx  cosx on the interval ,  has the form
Fx  sinx  C (where C can be any constant).
We want to find the particular antiderivative of f such that F0  6. To do this, we must
solve the equation
sin0  C  6.
Since sin0  0, the solution of the above equation is C  6. We conclude that the
particular antiderivative, F, of the function fx  cosx on the interval ,  that
satisfies the condition F0  6 is
Fx  sinx  6.
Example Given that
f  x  12x 2  6x  4 for all x  , .
and that f0  4 and f1  1, find the function f.
Example The graph of a function, f, with domain ,  is given below. Suppose that F is
the particular antiderivative of f on ,  that satisfies the condition F0  0. Sketch the
graph of F. (You may also assume that the graph of F has two horizontal asymptotes.)
Graph of f
Rectilinear Motion
Recall that if
s  ft
is the displacement function for an object moving along a straight path, then the velocity
function of the moving object is
v  ds  f  t,
dt
and the acceleration function is
2
a  dv  d 2s  f  t.
dt
dt
Thus, the velocity function is the derivative of the displacement function, and the acceleration
function is the derivative of the velocity function (or the second derivative of the displacement
function).
Looking at this in reverse fashion, we see that the velocity function is an antiderivative of
the acceleration function, and that the displacement function is an antiderivative of the velocity
function.
In studying the motion of free–falling objects (under simplified conditions including no air
resistance), we may assume that the only force acting on the falling object is the force of
gravity. It is known (from Physical experimentation) that the acceleration caused by gravity is
a constant and is equal to approximately 9. 8 m / s 2 or approximately 32 ft / s 2 .
Example A ball is thrown upward with a speed of 48 ft / s from the edge of a cliff that is
432 ft high. Assuming that only force of gravity acts in accelerating the ball, determine the
acceleration, velocity, and displacement functions of the ball. How long does it take for the
ball to reach the ground (below the cliff)?
Solution Since gravity is the only force accelerating the ball, the acceleration function of
the ball is
a  32.
Since the velocity function is an antiderivative of the acceleration function, the velocity
function has the form
v  32t  C
where C is some particular constant. To determine what C is we use the fact that
v| t0  48.
This gives us the equation
 320  C  48
from which we conclude that C  48. Therefore, the velocity function of the ball is
v  32t  48.
Since the displacement function is an antiderivative of the velocity function, the displacement
function has the form
s  16t 2  48t  C
where C is some particular constant. To determine what C is we use the fact that
s| t0  432.
(Note: We are taking the reference point of motion to be the ground below the cliff.)
Solving the equation
 160 2  480  C  432
gives us C  432. Therefore, the displacement function of the ball is
s  16t 2  48t  432.
To find out how long it takes for the ball to reach the ground, we need to set s  0
and solve for t. By using the quadratic formula, we see that the solutions of the equation
 16t 2  48t  432  0
are
3  3 13
t
 6. 9
2
and
3  3 13
t
 3. 9.
2
The negative value of t is not the one we are looking for (since t  0 is the time at which the
ball was first thrown into the air). We conclude that the ball reaches the ground (at the base
of the cliff) about 6. 9 seconds after it is thrown.