Lecture (1)
Antiderivatives and the Rules
of Integration
BY
DR. SAFA AHMED ELASKARY
FACULTY OF ALLIED MEDICAL OF
SCIENCES
A function F is an antiderivative of f on an interval I if
F’(x)=f(x) for all x in I.
Let’s use an example to figure out what this statement means…
Suppose we know f(x) = 2x and we want to find its antiderivative, F.
If f(x) = 2x, then F’(x) = 2x. So we know the derivative of F.
Think backwards, what function has a derivative equal to 2x ?
F(x) = x² !!!
To find the antiderivative, do the reverse of finding the derivative.
Is F(x) = x² the only function whose derivative is 2x ? Or in other
words, is F(x) = x² the only antiderivative of 2x ?
Theorem 1: Let G be an antiderivative of a function f. Then,
every antiderivative F of f must be of the form
F(x) = G(x) + C, where C is a constant.
So the answer to the question is F(x) = x²the only antiderivative of
2x is NO!!
Example 1: Let F(x) = x² + 4 and let G(x) = x²- 1
Then F’(x) = 2x and G’(x)= 2x
Thus both F and G are antiderivatives of f(x) = 2x.
Note two functions which have the same derivative will only
differ by the constant.
This means the antiderivative of a function is a family of
functions as pictured on the next slide.
The Indefinite Integral
The process of finding all antiderivatives of a function is
called antidifferentiation, or integration.

is the symbol used for integration and
is called the integral symbol.
We write
 f(x)dx  F(x)  C
This is called an indefinite integral, f(x) is called the
integrand and C is called the constant of integration.
Basic Integration Rules

Rule 1: kdx  kx  C (k, a constant)
Keep in mind that integration is the reverse of differentiation.
What function has a derivative k?
kx + C, where C is any constant.
Another way to check the rule is to differentiate the result and see
if it matches the integrand. Let’s practice.
Example 2:
 2dx  2x  C
Example 3:
2
2

dx


xC

Before we list Rule 2, let’s go back and think about derivatives.
When we used the power rule to take the derivative of a power, we
multiplied by the power and subtracted one from the exponent.
Example:
d 3
x  3x2
dx
 
Since the opposite of multiplying is dividing and the opposite of
subtracting is adding, to integrate we’d do the opposite. So, let’s try
adding 1 to the exponent and dividing by the new exponent.
31
x
1 4
Integrating: x3dx 

x

31 4
Check by differentiating the result:
d 1 4 4 3
3
 x  x x
dx  4  4
Since we get the integrand we know it works.
Basic Integration Rules
Rule 2: The Power Rule
xn 1
x  n 1  C
Example 4: Find the indefinite integral
Solution:


t 3 dt
t4
t dt 
C
4
3
Example 5: Find the indefinite integral
Solution:
n  1
n

3
x 2 dx
3
1
2
x
5
X2
x
3
2
dx
5
2

C 
C  X2 C
3
5
5
1
2
2
Here are more examples of Rule 1 and Rule 2.
1
Example 6: Find the indefinite integral  3 dx
x
1
Solution:
x
3

dx  x
3
x 31
x 2
1

C 
C  2 C
31
2
2x
Example 7: Find the indefinite integral
 1dx  x  C
Solution:
Example 8: Find the indefinite integral
Solution:
 1dx


 3x 2 dx  3 x 2
2

3
x
dx

 3x 21
 3x 1
3

C 
C  C
2 1
1
x
Basic Integration Rules
Rule 3: The Indefinite Integral of a Constant Multiple of a Function
 cf(x)dx  c f(x)dx, c is a constant
Rule 4: The Sum Rule (or difference)
 fx  gxdx   f(x)dx   g(x)dx
 fx  gxdx   f(x)dx   g(x)dx
Rule 5:
 e dx  e
Rule 6:
1
 xdx  ln x  C
x
x
C
To check these 2 rules, differentiate
the result and you’ll see that it matches
the integrand.
Example 9: Integrate.
 (2x 
1 3
 2  x  3e x )dx
x x
Using the sum rule we separate this into 5 problems.
1
3
x
2
xdx

dx

dx

x
dx

3
e

 x  x2 
 dx
Call them:
1
2
3
4
5
For 11 we will use rule 3 to bring the constant outside of the
integral sign.
2 xdx
Next we will use rule 2, the power rule to integrate.
 x11 
x2
  2
2 xdx  2
 x2
2
1 1 
Example 9 continues…
1
3
x
2
xdx

dx

dx

x
dx

3
e

 x  x2 
 dx
Call them:
1
For
2
1
2
3
4
5
2
2
xdx

x

we will use Rule 6 the natural log rule.
1
  dx   ln x
x
For 3 we will first rewrite then use the constant rule (Rule 3)
and then the power rule (Rule 2).
21
1
3
x
x
3
2
 x2 dx  3 x dx  3  2  1  3  1  x
Example 9 continues…
1
3
x
2
xdx

dx

dx

x
dx

3
e

 x  x2 
 dx
Call them:
1
 2xdx  x
2
1
2
2
3
4
1

dx  ln x
x

5
3
3
x
dx 
2
3
x
For 4 we will rewrite and then use the power rule (Rule 2).

1
2
1
1
2
3
x
2
xdx   x dx 
 x2
1
1 3
2
For 55 we will use the constant rule (Rule 3) and then Rule 5 for ex.
x
x
x
3
e
dx

3
e
dx

3
e


Example 9 continues…
1
3
x
2
xdx

dx

dx

x
dx

3
e

 x  x2 
 dx
Call them:
11
 2xdx
x
2
2
3
44
3
1
  dx   x 2 dx
x
3
 ln x

x
  xdx
2 23
 x
3
5
  3e x dx
 3e x
So in conclusion:
3
1
3
3 2 2
x
2
x
2
xdx

dx

dx

x
dx

3
e
dx

x

ln
x


x

3
e
C

 x  x2 

x 3
You may be wondering why we didn’t use the C before now. Let’s
say that we had five constants C1  C2  C3  C4  C5 . Now we add all
of them together and call them C. In essence that’s what’s going
on above.
Here are some for you to try:
1.
Integrate and check your answer by taking the derivative.

3  x 4  e x dx
Click the correct answer below.
3x  5x 5  e x  C
3x  4x 3  e x  C
3x 
1 5
x  ex  C
5
Sorry that’s not correct.
Think about the power rule for integration. You should add one to
the exponent and divide by the new exponent.
Try again. Return to the previous slide.
Good Work!!
Here is the solution in detail.



3  x 4  e x dx  3dx  x 4 dx 

e x dx
x 4 1
 3x 
 ex  C
4 1
1 5
 3x  x  e x  C
5
2. Integrate.

2x  4 
1
 x2

3
dx
x
Click on the correct first step.
Integrate each factor separately .
x
2

 1

 4x 
 3ln x   C
 x

Multiply the two factors in the radicand together
and combine the like terms.

6
10
4
 2 dx
x x
Be careful !!
Rule 4 states:
 fx  gxdx   f(x)dx   g(x)dx
 fx  gxdx   f(x)dx   g(x)dx
This does NOT apply to multiplication or division.
You should multiply the factors in the integrand, simplify and then
use Rule 4 to integrate the terms.
Multiply the two factors in the radicand together
and combine the like terms.

2
4 12
6 2 
dx 
x
x
x

6
10
4
 2 dx
x x
Next use Rule 4 to integrate each term separately .

6dx  10

1
dx  4
x

x 1
6x  10ln x  4
C
1
6x  10ln x 
4
C
x
x 2 dx
Let’s look at the solution to the problem:
f’(x)= 3x2 - 4x + 8 

f(1)= 9

Solution: First integrate both sides:
3x3 4x2
f(x) 

 8x  C
3
2
Simplify:
f(x)  x3  2x2  8x  C
Now find C by using the initial condition.
Substitute 1 for x and 9 for f(x)
9  1  21  81  C
9  1 28 C
9 7 C
2C
3
2
This gives the particular
solution.
f(x)  x3  2x2  8x  2
Review - Basic Integration Rules

Rule 1: kdx  kx  C (k, a constant)
Rule 2: The Power Rule
xn 1
x  n 1  C
n
Rule 3: The Indefinite Integral of a Constant Multiple of a Function
 cf(x)dx  c f(x)dx, c is a constant
Rule 4: The Sum Rule (or difference)
 fx  gxdx   f(x)dx   g(x)dx
 fx  gxdx   f(x)dx   g(x)dx
x
x
e
dx

e
C
Rule 5: 
1
Rule 6:  xdx  ln x  C