Download Anti-Derivatives

4.2 Revisited
Anti-Derivatives
y
C
These two functions have the
same slope at any value of x.
Functions with the same derivative
differ by a constant.
y  g  x
x
0
y  f  x

Instead of taking a function and finding it’s derivative,
let’s take a known derivative and see if we can back track
to the original function.
Find the function f  x  whose derivative is sin  x  and
whose graph passes through  0, 2  .
In other words, let’s do things backwards. We have f
and we want to find f.
d
cos  x    sin  x 
dx
 f  x    cos  x   C
d
so:
 cos  x   sin  x 
dx
Wait! Where did the C come from?

g ( x)  x  5
f ( x)  x  2
2
2




These two functions
only differ by a
constant, but they
have the same
derivative.
f ( x)  g ( x)  2 x















So working backwards from 2x, we have to allow for a
constant C because we can’t know the exact value of C
until we know more about the original function.
Instead of taking a function and finding it’s derivative,
let’s take a known derivative and see if we can back track
to the original function.
Find the function f  x  whose derivative is sin  x  and
whose graph passes through  0, 2  .
In other words, let’s do things backwards. We have f
and we want to find f.
d
cos  x    sin  x 
dx
 f  x    cos  x   C
d
so:
 cos  x   sin  x 
dx
f  x  could be  cos  x  by itself or could vary by some
constant C .
Instead of taking a function and finding it’s derivative,
let’s take a known derivative and see if we can back track
to the original function.
Find the function f  x  whose derivative is sin  x  and
whose graph passes through  0, 2  .
In other words, let’s do things backwards. We have f
and we want to find f.
Plug in (0, 2)
d
cos  x    sin  x 
dx
d
so:
 cos  x   sin  x 
dx
 f  x    cos  x   C
2   cos  0  C
2  1  C
Notice that we had to have initial
values to determine the value of C.
3C
f  x    cos  x   3

The process of finding the original function from the
derivative is so important that it has a name:
Antiderivative
A function F  x  is an antiderivative of a function f  x 
if F   x   f  x  for all x in the domain of f. The process
of finding an antiderivative is antidifferentiation.
You will hear much more about antiderivatives in the future.
This section is just an introduction.

Many anti-derivatives like sin(x) and cos(x) are intuitive.
For polynomials, we need a power rule as we did for
derivatives:
Power Rule for
derivatives:
Power Rule for
anti-derivatives:
1. Multiply by the
exponent
2. Subtract 1 from the
exponent.
d n
x  nx n 1
dx
n 1
x
xn 
C
n 1
Eventually, you will have to
use the notation for antiderivatives which looks like
this:
1. Add 1 to the exponent
2. Divide by the new
exponent
n 1
x
 x dx  n  1  C
n

Example 7b: Find the velocity and position equations
for a downward acceleration of 9.8 m/sec2 and an
initial velocity of 1 m/sec downward.
a (t )  9.8
v(t )  9.8t  C
 1  9.8(0)  C
1  C
v(t )  9.8t  1
Velocity is the anti-derivative
of acceleration
Example 7b: Find the velocity and position equations
for a downward acceleration of 9.8 m/sec2 and an
initial velocity of 1 m/sec downward.
a (t )  9.8
v(t )  9.8t  C
 1  9.8(0)  C
1  C
v(t )  9.8t  1
 9.8 2
s (t ) 
t t C
2
The power rule in reverse:
Increase the exponent by one and
multiply by the reciprocal of the
new exponent.
Since velocity is the derivative of position,
position must be the antiderivative of velocity.

Example 7b: Find the velocity and position equations
for a downward acceleration of 9.8 m/sec2 and an
initial velocity of 1 m/sec downward.
a (t )  9.8
v(t )  9.8t  C
 9.8 2
s (t ) 
t t C
2
s(t )  4.9t  t  C
2
 1  9.8(0)  C
The initial position is zero at time zero.
1  C
v(t )  9.8t  1
0  4.9(0)  0  C
2
0C
2
s(t )  4.9t  t

The process of finding the original function from the
derivative is so important that it has a name:
Antiderivative
A function F  x  is an antiderivative of a function f  x 
if F   x   f  x  for all x in the domain of f. The process
of finding an antiderivative is antidifferentiation.
Problems like these are called Initial Value Problems
because an initial value is needed to find C.
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