Transcript 13.1 Antiderivatives and Indefinite Integrals

13.1 Antiderivatives and
Indefinite Integrals
The Antiderivative
The reverse operation of finding a derivative is called the
antiderivative. A function F is an antiderivative of a function f if
F ’(x) = f (x).
1) Find the antiderivative of f(x) = 5
Theorem 1:
Find several functions that have the derivative of 5
If a function
has more than
one
antiderivative,
then the
antiderivatives
differ by at
most a
constant.
Answer: 5x; 5x+ 1; 5x -3;
2) Find the antiderivative of f(x) = x2
Find several functions that have the derivative of x2
1 3
1 3
1 3
x ; or x  ; or x  e
Answer:
3
3
3
• The graphs of antiderivatives are vertical
translations of each other.
• For example: f(x) = 2x
Find several functions that are the antiderivatives
for f(x)
Answer: x2,
x2 + 1,
x2 + 3,
x2 - 2,
x2 + c (c is any real number)
Indefinite Integrals
Let f (x) be a function. The family of all functions that are
antiderivatives of f (x) is called the indefinite integral and has
the symbol f ( x) dx

The symbol  is called an integral sign, and the function f (x) is
called the integrand. The symbol dx indicates that antidifferentiation is performed with respect to the variable x.
By the previous theorem, if F(x) is any antiderivative of f, then
 f ( x) dx  F ( x)  C
The arbitrary constant C is called the constant of integration.
Indefinite Integral
Formulas and Properties
n 1
x
1.  x n dx 
 C , n   1 (power rule)
n 1
2.
x
x
e
dx

e
C

3.
1
 x dx  ln | x | C
4.
5.
 k f ( x) dx  k  f ( x) dx
  f ( x)  g ( x) dx   f ( x) dx   g ( x) dx
It is important to note that property 4 states that a constant
factor can be moved across an integral sign. A variable
factor cannot be moved across an integral sign.
Example 1:
A)
 2dx  2 x  C
B)
 16e dt  16e
C)
t
 3x dx 
4
t
C
 x5 
3 5
3   C  x  C
5
 5 
Example 1 (continue)
D)
 (2 x
5
 3 x  1)dx 
2
  2 x dx   3 x dx   1dx
5
2
 x6   x3 
 2   3   1x  C
6 3
1 6
3
 x  x  xC
3
Example 1 (continue)
E)
5
x

4
e
dx



 x

5
x
  dx   4e dx
x
1
x
 5 dx  4 e dx
x
 5 ln x  4e  C
x
Example 2
A)
 23 3 


2
x

dx 
4 
 
x 
2
3
2
3
  2x dx   3x 4dx  2 x dx  3 x 4dx


 5
5
 x 3   x 3 
6 3
3
  C  x  x  C
 2   3
5
 5   3 
 
 3 
5
6 3 1
 x  3 C
5
x
Example 2 (continue)
B)
4
5
3
5
w dw  4 w dw 

3
 8
 w5
 4
 8

 5

8

5
 C  x5  C

2


Example 2 (continue)
 x  8x

  x 2
4
C)
3

dx 

 x
2

 8 x dx 
  x dx   8 xdx
2
2

x
x 

 8   C
3
 2 
3
x
2
  4x  C
3
3
Example 2 (continue)
D)
1
3
1

2
6 
 3
  8 x  x dx  8 x dx  6 x dx 

 4
 x3
 8
 4

 3
  1
  x2
  6
  1
 
  2
4
3



C



 6 x  12 x  C
Example 2 (continue)
E)
 ( x  2)( x  3)dx 
2
 x
3

 3 x  2 x  6 dx 
2
  x dx   3 x dx   2 xdx   6dx
3
2
x 4  x3   x 2 
  3   2   6 x  C
4
3  2
4
x

 x3  x 2  6 x  C
4
Example 3
Find the equation of the curve that passes through (2,6) if its
slope is given by dy/dx = 3x2 at any point x.
The curve that has the derivative of 3x2 is
 3x dx 
2
 x3 
3   C  x 3  C
 3
Since we know that the curve passes through (2, 6), we can find out C
3
y  x C
6  23  C
6 8C
C  2
Therefore, the equation is y = x3 - 2
Example 4
Find the revenue function R(x) when the marginal revenue is
R’(x) = 400 - .4x and no revenue results at a 0 production
level. What is the revenue at a production of 1000 units?
The marginal revenue is the derivative of the function so to find the
revenue function, we need to find the antiderivative of that function
 (400  .4 x)dx  
 x2 
400 dx  .4 xdx  400x  .4   C  400x  .2 x 2
 2

So R(x) = 400x -.2x2, we know need to find R(1000)
R(1000)  400(1000)  .2(1000)  200,000
2
Therefore, the revenue at a production level of 1000 units is $200,000
Example 5
The current monthly circulation of the magazine is 640,000 copies. Due
to the competition from a new magazine, the monthly circulation is
expected to decrease at a rate of C’(t)= -6000t1/3 copies per month, t is
the # of months. How long will it take the circulation of the magazine to
decrease to 460,000 copies per month?
We must solve this equation: C(t) = 460,000 with C(0) = 640,000
To find the function C(t), take the antiderivative
 4
t3
 6000t dt   6000 4


3
 43 


C (0)  4500 0   C
 

1
3
 43 
640,000  4500 0   C
 
640,000  C
4
3
C (t )  4500t  640,000

4

  C  4500t 3  C



4
3
460,000  4500t  640,000
 180,000  4500t
40  t
4
3

  t

15.9  t
40
3
4
4
3




3
4
So, it takes about 16 months
4
3