```Section 7.1 – Antiderivatives
d 3
x  = 3x 2

dx
Recall:
F(x)
3x 2 and find the function, F, whose derivative is f(x) = 3x2 .
f(x)
d 3
x   3x 2

dx
Antiderivative
Derivative
The function F is called an antiderivative of f.
The process of finding a function F(x) from its derivative f(x) is called antidifferentiation.
Example: Find an antiderivative of f(x) = x2 .
Think: Find a function whose derivative is x 2 (UNDO differentiation).
Example: Find an antiderivative of f(x) = 5x 4 .
Think: Find a function whose derivative is 5x 4
1
Recall that the notation for taking a derivative is:
d
dx
The most traditionally used notation for finding an antiderivative is:
  function dx
Indefinite Integral (the most general antiderivative)
The set of all antiderivatives of the function f is the indefinite integral of f with respect to x,
denoted by
 f ( x)dx
The symbol

is an integral sign. The function f is the integrand of the integral, and x is the
variable of integration.
 f ( x)dx  F ( x)
means F ( x)  f ( x)
Examples: Find an antiderivative.
(a) f  x  = 6
(b) f  x  = x
(c) f  x  = 5x4
(d) f  x  = ex
(e) f  x  =
1
x
But, antiderivatives are NOT unique……
2
Consider:
d 2
x  =
dx
d 2
x  4 =
dx
d 2
 x  2
dx
Rules for Integration
(Memorize these!!!)
xn+1
 x dx = n + 1 + C
n
 kf(x)dx
"power rule"
= k  f(x)dx
  f(x) ± g(x) dx
=
x
x
 e dx = e + C
"pull constants to the front "
 f(x)dx ±  g(x)dx
and
"integrate separately"
kx
 e dx =
ekx
k
+C
ax
 a dx = lna + C
x
1
 x dx =
x
1
dx = ln x + C
3
Examples:
Find the indefinite integral for each of the following functions; check your answer by differentiation.
(a)
 2x dx =
(b)
 4x
(c)
 3e
(d)
5
 x dx =
(f)
 x  3 dx =
(g)
  2  dx 
3
x
dx =
dx =
x
4
(h)

(i)
 (8x
3
(j)
 e
+ x 2  12x  dx 
1
2
x 2  2x + 6  dx =
x
+ 4x 1 )dx 
5
(k)
1
1 

  y3 y  dy 


(l)
 x2 +1 
  x  dx 
6
Example: The marginal profit in dollars on Brie cheese sold at a market is given by:
P '(x)  50 x 3  30 x 2 where x is the amount of cheese sold, in hundreds of pounds.
When no cheese is sold, the profit is –\$40.
(a) Find the profit function.
(b) Find the profit from selling 200 lb of Brie.
7
Section 7.2 – Substitution
Now we will integrate when the Chain Rule is involved…...
The Substitution Method:
In general (for the types of problems we are concerned with) there are three cases.
We choose u to be one of the following:
1. The quantity under a root or raised to a power
2. the exponent on e
3. the quantity in the denominator
Consider the following:
4
3
2x
+1
8x
dx



5
x
2xe
 dx
Let
then
du =
Let
u=
then
du =
u=
2
The Substitution Rule:
If g ( x ) is a differentiable function whose range is an interval I and f is continuous on I , then
 f  g(x) g(x) dx =  f(u)du = f(x) + C
Let u  g(x) and du  g '(x)dx .
Then integrate f(u).
In other words, g’(x) MUST be present if you are going to integrate f.
8
Examples: Use substitution to find each indefinite integral.
(a)
 6x (3x
(b)
2 x
3x
 e dx 
2
+ 4)4 dx 
3
(c)
(d)
 2x

1 + x2 dx =
3dt
=
3t  5
9
(e)

2x
6x 2
3
+ 7
3
2
dx =
2 + lnx
dx =
x
(f)

(g)
1
 x lnx  dx =
10
Now, what do you do if “du” is missing a constant…..?
(h)
 2x +1
(i)
 5e
(j)
3 x
 x e dx =
3
3x
dx =
dx =
4
11
(k)
(l)
2
3x
  x  1 e
3
9x
dx =
t2 + 2
 t3 + 6t + 3 dt =
12
Section 7.3 – Area and the Definite Integral
Areas can be approximated using the area of rectangles drawn by three different methods:
Left endpoints
Right endpoints
Midpoints
Example:
Use LEFT ENDPOINTS to approximate the area between the graph of f ( x ) 
y
y
y = 1/x
1
and the x-axis
x
from x = 1 to x = 3 . Use n = 2 and then use n = 4.






x













13
Example:
Use RIGHT ENDPOINTS to approximate the area between the graph of f ( x ) 
y
y
y = 1/x
1
and the x-axis
x
from x = 1 to x = 3 . Use n = 2 and then use n = 4.






x













Example:
What is the area between the graph of f(x) = x2 + 5x  4 and the x-axis?
Use three rectangles.
(a) Left Endpoints
y = -x^2+5x-4y


x






14


(b) Right Endpoints
y = -x^2+5x-4y


x








(c) Midpoints
y = -x^2+5x-4y


x








15
To develop a process that would yield the exact area, begin by dividing the interval from a to b into
n pieces of equal width, using each of these n pieces as the base of a rectangle. Let x1 be an
arbitrary point in the first interval, x2 be an arbitrary point in the second interval, and so on, up to
the nth interval.
The symbol x is used to represent the width of each of the intervals.
Area of the ith rectangle = f xi   x
n
Area of all n rectangle =
 f  x   x
i 1
i
n
Exact area = lim
 f xi   x
n 
i 1
*Called a Reimann Sum
The indefinite integral is a set of functions, however, the definite integral is a number.
We will find the exact areas using definite integrals in the next section (7.4)
16
Section 7.4 – The Fundamental Theorem of Calculus
Examples: Find the definite integrals.
2
(a)
 4t
3
dt =
1
  4z + 3 dz =
2
(b)
-2
17
(c)
(d)


2
1
9
3
dy
=
y
2x  2 dx =
18
4
(e)
3
 2p + 1
2
dp 
1
Example: A worker new to a job will improve his efficiency with time so that it takes him fewer
hours to produce an item with each day on the job, up to a certain point. Suppose the rate of
change of the number of hours it takes a worker in a certain factory to produce the xth item is given
by H'  x  = 20  2x . What is the total number of hours required to produce the first 5 items?
19
Area
In the previous section we saw, that, if f  x   0 in a, b , the definite integral
 f  x  dx
b
a
gives the area below the graph of the function f(x), above the x-axis, and between the
lines x  a , x  b .
How about if f(x) is negative?
If a function takes both positive and negative values, then the sum is the sum of the area of the
rectangles that lie above x-axis and the negatives of the areas of rectangles that lie below the x-axis.
b
 f ( x)dx  A
1
 A2
This is called the NET AREA.
a
Where A 1 = the area of the region above the x-axis
A 2 = the area below the x-axis
Example: Find the area between the x-axis and the graph of f ( x)  x 2  4 from x  0 to x  2
















20
Example: Use the definite integral to find the area between the x-axis and f (x) over the indicated
interval.

(a) f  x  = 9  x ;  0, 6
2




















x
(b) f  x  = e  1;  1, 2









21
Section 8.2 – Average Value
What does it mean to find an “average”?
Example: Find the average value of the function f  x  =  2x  1  2 on the interval 1, 13 .
1
22
Example: A company plots the price per share of a stock that is owned as a function of time and
finds that it can be approximated by the function S(t )  25t  5t 2  18 where t is the time (in years)
since the stock was purchased. Find the average price of the stock over the first five years.
23
Section 10.1 – Solutions of Elementary and
Separable Differential Equations
The solution of a differential equation is a set of functions.
Case I: Elementary Differential equations in the form:
Consider the equation
dy
= f(x)
dx
dy
= 3x 2  2x consists of all expressions for y(x) that have a derivative
dx
equal to 3x 2  2x .
y    3x 2  2x  dx = x 3  x 2 + C
Each different value of C leads to a different solution of the
equation, showing that a differential equation can have
an infinite number of solutions.
Examples: Find the general solution for each differential equation.
(a)
dy 2
= x 1
dx
(b)
dy
= 3e 2x
dx
24
Case II: Differential Equations in the form
Think of
dy f(x)
=
dx g(y)
(called separable DE)
dy
as a fraction of dy over dx.
dx
Multiply on both sides to get:
dy f(x)
=
 g(y)dy = f(x)dx
dx g(y)
Notice the left side of the equation is in terms of y, and the right side is in terms of x.
We have separated the variables.
Now take the antiderivatives of f and g.
 g(y)dy =  f(x)dx
 G(y) = F(x) + C
Examples: Solve the following by separation of variables.
(a)
dy x2
=
dx y
(b)
dy
= y5
dx
25
(c)
dy ex
=
dx e y
(d)
dy
= y 2e x
dx
26
An Initial Value Problem is a differential equation with a value of y given for some
value of x.
(you will use the given x and y to solve for C)
Examples: Find the particular solution for each initial value problem.
(a)
dy
 2x  3x 2
dx
(b)
dy 3x + 1
=
dx y  9
y(0) = 5
y(0) = 2
27
Example: Find the general solution of the differential equation
So,
dA
= kA
dt
dA
= kA
dt
 A = Mekt
Example: Sales (in thousands) of a certain product are declining at a rate proportional to the
amount of sales, with a decay constant of 15% per year.
(a) Write a differential equation to express the rate of sales decline.
(b) Find a general solution to the equation in part a.
28
(c) Suppose that when t = 0, sales are \$2,000. Find the particular solution.
(d) What will sales be after 6 years?
29
```