Topics In Integral Calculus
260 BC Archimedes found the area of a circle, the volume of a
sphere, the surface area of a sphere and the area
under the graph of a parabola .
1675 Leibnitz used ∫f(x) dx for the “reverse operation” of df(x)
and gave the first method to find the reverse of the
combination of functions that were known at that time. This
work was later completed by Euler in the XVIII century.
1823 Cauchy provided the first rigorous treatment of integration.
1830
Riemann gave a more theoretical definition of the defined
integral for continuous functions.
ANTIDERIVATIVE OF A CONTINUOUS FUNCTION
We say that F(x) is an antiderivative (the reverse process of
differentiation) of f(x) on an interval I when F’(x) = f(x) for all x in I.
For example, x2 is an antiderivative of 2x, since (x2)’ = 2x. Of
course x2+C is an anti-derivative of 2x for any real number C.
We use the notation,∫ f(x) dx, to represent the set of all antiderivatives
of f(x). Thus we write: ∫ 2x dx = x2 + C , where C is any real number
Ex 1: Find ∫ x3 dx
We need to find an antiderivative for x3 .
Since ¼ (x4 )’=4 x3  (¼ x4 )’= x3. So ∫ x3 dx = ¼ x4 + C
The general integration formula for power functions will be listed on next
slide.
Ex 2: ∫sin x dx = - cos x + C , since (- cos x + C)’ = sinx
Remark: Let F(x) and G(x) be any two antiderivatives of f(x).
Then [ G(x) – F(x)] ’ = G’(x) – F’(x) = f(x) – f(x) = 0.
So G(x) - F(x) = C and from here we get G(x) = F(x) + C, this says
that any two antiderivatives of f(x) differ only by a constant.
Teorem: ∫ f(x) dx = F(x) + C , where F(x) is any antiderivative of f(x).
For every known derivative formula, when reading backward, it
will give us back its own anti-derivative. For example:
(sin x)’ = cos x
(cos x)’ = - sinx
(tan x)’ = sec2x
(sec x )’ = sec x tanx
(csc x )’ = - csc x cot x
(cot x )’ = - csc2 x
(ex ) ’ = ex
(ln x)’ = 1/x
(sin-1 x)’ = (1- x2) -1/2









∫ cos x dx = sin x +C
∫ sinx dx = - cos x +C
∫ sec2x dx = tan x + C
∫ sec x tan x dx = sec x + C
∫ csc x cot x dx = - csc x + C
∫ csc 2 x dx = - cot x + C
∫ ex dx = ex + C
∫ 1/x dx = ln x + C , for x>0
∫ (1- x2) -1/2 dx = sin-1 x + C
(tan-1 x)’ = (1+x2) -1

∫ (1 + x2) -1 dx = tan-1 x + C
Remark:
1) If we change 3 in Ex 1 for any number n   1 , we get
. ∫ x n dx = x n+1 /(n+1) + C , since (xn+1 /(n+1) )’ = (n+1) x n /(n+1) = x n
2) ∫k f(x)dx = k ∫f(x)dx , since [k ∫f(x)dx ] ’ = k [∫f(x)dx ]’ = k f(x)
3) ∫[f(x)  g(x) ] dx = ∫f(x)dx  ∫g(x)dx , since [∫[f(x)  g(x) ] dx ]’ = f(x)  g(x)
Ex 3 : Find
∫(8x3 - 10x + 4) dx
= ∫(8x3 dx -∫10x dx + ∫4 dx
Using 2) … = 8∫x3 dx - 10∫x dx + 4∫dx
Using 1) … = 8 x4/4 - 10x2/2 + 4x +C
Simplifying… = 2x4 - 5x2 + 4x +C
Using 3) …
Ex 4 : Find
∫(1- 1/x4 + cos x) dx
= ∫dx -∫x -4 dx + ∫cos x dx
= x - x -4+1 /(-4+1) + sinx +C
= x +(1/3)x -3 +sin x + C
The following examples show us that using antiderivative to
recover quantities from their derivatives.
Ex5: Find the equation of the curve that has a tangent line
at any point (x, y) with slope 4 x  3 and passing through
the point (2, 5).
2
Sol:
y   (4 x  3) dx  x  3 x  C
 (4 x  3) dx  2 x  3 x  C  y
2
A t the point (2, 5) : 8  6  C  5  C  3
T hus y  2 x  3 x  3 is the required curve.
2
Ex 6: A company find that the marginal cost when x un its are
produced is C ( x )  (0.02 x  40) dollars. If the fixed cost is $1000.
Find the cost function and the cost produced 50 units.
C (x) 
 (0.02 x  40) dx  0.01 x  40 x  C
2
A t C (0)  1000 : C  1000  C ( x )  0.01 x  40 x  1000
2
A t x  50 : C (50)  25  2000  1000  3050
Ex 7: Knowing that f ’’(x) = 12x- 4 , f ’(2)=10 & f(5)=200, find f(x)
f ’(x) = ∫ f ’’(x) dx =∫(12x-4)dx = 6x2 – 4x+C. But f ’(2)= 10.
So 10 =6(4)-4(2)+C  C=-6  f ’(x) = 6x2 – 4x-6
Then f(x) = ∫(6x2 – 4x-6 ) dx = 2x 3 – 2x 2 – 6x + K
But f(5)= 200 , so 200= 2( 5 3 ) – 2(5) 2 – 6(5) + K  K = 30
Finally f(x) = 2x 3 – 2x 2 + 6x +30
Ex 8: A car is traveling on a high way (straight line). Its velocity
function is given by v(t)= 3t2-2t+65 mph. When t=3 hr. the car was
220 mi from the starting point (origin). Find the position function of the
car.
2
3
2
Sol: S ( t )   v ( t ) dt    3 t  2 t  65 dt  t  t  65 t  C
220  S (3)  27  9  195  C  C  7  S ( t )  t  t  65 t  7
3
2
Ex 9: A car is traveling on a long straight high way at a velocity v(t) in
mph is given by the formula v(t) = 100-50t. When t= 0 the position of
the car is located at100 miles from the origin. Find the position of the
car located after 2,4 &5 hours? Explain.
2
v(t)
s
(
t
)

v
(
t
)
dx

(

50
t

100)
dt


25
t

100
t

K
Since


100
Using that s(0)=100 we get that K=100
So , s ( t )   25 t  100 t  100
2
Replacing t for 2,4 & 5 we get:
s (2)   25(2)  100(2)  100  200
2
s (4)   25(4)  100(4)  100  100
2
s (5)   25(5)  100(5)  100   25
2
t =0
When v(t)=0  t=2 and the velocity is negative for
t>2, which means that the car is turning around
toward the origin when t > 2 and continuing ….
4
2
t
Find the area under the curve f(x) = x2
Y for 0x4.
16
1) Let h=(b-a)/n = (4-0)/4 =1, and
use 4 rectangles, each with base h
and height f(x) evaluated at the left
end point of each subinterval.
2) Let h=(4-0)/8 =½ and use 8
rectangles, each with base h and
height f(x) evaluated at the left end
point of each subinterval.
3) Let h=(4-0)/16 =¼ and use 16
rectangles, each with base h and
height f(x) evaluated at the left end
point of each subinterval.
9
4) Let h=(4-0)/n and as n ∞, it is
easy to “see” that the value of this
limit is “the value of the area under
the parabola y=x2 for 0x4
4
3
2
1
¼
½
1
2
3
4
X
Remark: The problem of this method is finding the unwieldy sum Sn .
Arhimedes (400 BC) used a similar method to find the area under a
simple curve like a parabola.
Now we’ll use other approach to find area.
Let y=f(x) be a positive continuous function defined on [a,b]
First, let’s see the area in a Kinematics way (sliding doors effect …)
and call A(x) = the area under y=f(x) from x=a to any variable x.
Then
satisfies: 1) A(a)=0 , and 2) A(b) = Area
It’s clear that A(x) is an increasing continuous function. Let’s find its derivative.
We’ll try to find the bound for
[m in f ( x )]  x
x

A( x  x)  A( x)
x
A( x  x)  A( x)
x

[m ax f ( x )]  x
.
y=f(x)
Y
x
Taking the limit when x0 and we see that
A(x)
a
x
we then get that A’(x) = f(x)
x+x
b
X
Min f (x) = f (x) = Max f (x)
Let  x 
40

n
4
The right end point of each subinterval is
.
n
4
xi  a  i x  0  i  
n
The sum of the area of these n rectangles under the curve
f(x) = x2 on [0,4] is given by:
n
Sn 

i 1
n
 4   4  64
2
f ( xi )   x    xi    x    i     3  i
n i 1
i 1
i 1  n   n 
n
n
2
2
Now we use a well known formula for the sum of the first n
squared integers.
n
i
2

n ( n  1)( 2 n  1)
6
i0
So Area = lim S n  lim
n 

32
3
n 
(2) 
64
3
64 n ( n  1)( 2 n  1)
6n
3
 lim
n 
32 
1 
1  
3 
n 

2 
1
n

So A(x) is an anti-derivative of f(x), that satisfies: A(a)=0 & A(b)=Area
Let F(x) be any anti-derivative of f(x) then: A(x) = F(x) + C
So A(a) = F(a) + C  0 = F(a) + C = 0  C = - F(a)
A(b) = F(b) + C  Area under f(x) on [a,b] = F(b)-F(a)
Remark : This means that we can use any anti-derivative to
calculate the area. We’ll use the following notation:
b
A rea 

Y
f ( x ) dx  F ( b )  F ( b )   F ( x )  a
b
f ( x)  x  2 x
2
8
a
Ex 10: Find the area of the region bounded by
f ( x)  x  2 x , y  0 & x  2
2
The shaded area under the curve:
2
x

2
A rea   ( x  2 x ) dx  
x 
0
3
0
3
3
3
2
2
2
0
8
20
2 
2 
  2  0   4
3
3
 3
 3
X
-2
0
2
Remark: It’s clear that for f(x) positive over [a,b] we get:

b
n
f ( x ) dx  lim
n 
a
If f(x) is negative

b

f ( x i )  x  A rea under f ( x ) over [ a , b ]
i 1
f ( x ) d x will give the negative of the area
below [a,b]
a
For this reason we are going to extend the meaning of

b
f ( x ) dx to
any .real values of a & b , and f(x) being any continuous function. And
be call it definite integral from a to b.
a
Ex 11 : Let f(x) be the function defined by the graph on the right. Find
y=f(x)
0
4 Y
---------------------------------a )  f ( x)dx 
½(2)(4)= 4
2
4
b )  f ( x ) d x  -½(4)(4)= - 8
0
c)
d)

6
f ( x)dx 
½(2)(4)= 4
4

6
2
f ( x)dx 
X
-2
4
4-8+4 = 0
-4 -------
6
Fundamental Theorem of Calculus: Let f(x) be a continuous function
and F(x) be any anti-derivative of f(x), then

b
f ( x ) dx  F ( x )] a  F ( b )  F ( a )
b
a
The proof is similar to the one on slide #8.
Ex 12: Find I 

3x  2x  3x  3
3
3
2
1
Dividing …
I 

dx
x
3
(3 x  2 x  3 
2
1

3
x
) dx
 x  x  3 x  3 ln | x | ]1
 
3
2
3

 3  3  3(3)  3 ln 3  1  1  3(1)  ln 1  28  3 ln 3
Ex 13: Find I 
Dividing … I 

64
3
1
1
 x
64
1
2
3
1 / 3
x
3
2
dx
x
x
1/ 6

 3x
dx  
 2
2/3
 48 768   3 6  45 762 1839



   
7  2 7
2
7
14
 2


6x
7/6
7
64


1
Ex 14: Suppose the rate change of temperature of a cup of tea is
observed to be T  ( t )  15 e  0.3 t in Fahrenheit degree per minute
where t is time in minutes. What is the total change in temperature
of the cup of tea from t = 1 minute to t = 10 minutes.
Sol:
The total change in the temperature from t = 1 min to t = 10 min:
10

10
1
15 e
 0.3 t
 1  0.3 t 
0
dt  15  
e

34.55
F

0.3

1