THE AREA PROBLEM
tangent lines
Differential calculus
rates of change
Calculus (Fundamental Theorem of Calculus)
Area
Method of exhaustion (Archimedes 287 BC-212BC)
(rectangle method – sum of areas)
Antiderivative method - A( x)  f ( x) (derivative of
area function)
Integral calculus
Volume
definite integral
Antidifferentiation (Integration) – finding the
function when the derivative is known
antidifferentiation
Integration – The Area Problem
f (x )
A  A(x )
a
Area Problem
b
Rectangular Method
 Divide the interval into n subintervals
 As n gets bigger, the approximation gets better
 A(x)   intervals
1
Example 1:
y  x2
Find the area between 0 and 1.
If n = 4, then x 
1 1 3
, , ,1 so…
4 2 4
Now do n  10,100,1000,......
*GDC 2nd STAT  MATH, Sum, 2nd STAT OPS, seq(
 sum(seq( x 2 , x,0,1, 1 )) =33.835 = length
100
then 33.835 
1
 0.33835
100
Note the difference between using the left and right endpoints of the intervals (lower and
upper Riemann sums – see below). The lower and upper depend on whether the function is
increasing or decreasing over the given interval.
The Riemann Sum
Using n intervals of equal width between x =a and x=b, the width of each interval, or Δx :
𝑏−𝑎
𝑏−𝑎
∆𝑥 = 𝑛
Therefore the area of each interval is ( 𝑛 ) (𝑓(𝑥𝑘 ), where k indicates each
successive x-value, starting at 𝑘 = 1 and ending at 𝑘 = 𝑛.
𝑏−𝑎
𝑏−𝑎
𝑏−𝑎
If we add up all of these areas: ( 𝑛 ) (𝑓(𝑥1 ) + ( 𝑛 ) (𝑓(𝑥2 ) + ⋯ + (
approximation of the area under the curve. In sigma notation, this is:
(
𝑏−𝑎
𝑛
𝑏−𝑎
) (𝑓(𝑥1 ) + (
𝑛
𝑏−𝑎
) (𝑓(𝑥2 ) + ⋯ + (
𝑛
) (𝑓(𝑥𝑛 ) = ∑𝑛𝑘=1 (
𝑏−𝑎
𝑛
𝑛
) (𝑓(𝑥𝑛 ) we get an
) (𝑓(𝑥𝑘 ))
or ∑𝑛𝑘=1(𝑓(𝑥𝑘 )∆𝑥).
This is something that will be very important in first year calculus.
Left endpoint: ∑𝑛𝑘=1(𝑓(𝑥𝑘−1 )∆𝑥)
Right endpoint: ∑𝑛𝑘=1(𝑓(𝑥𝑘 )∆𝑥)
2
However – an important result we will be using is that if we take the limit as 𝑛 → ∞, we
have:
𝑏
lim ∑𝑛𝑘=1(𝑓(𝑥𝑘 )∆𝑥) which is approximately equal to: ∫𝑎 𝑓(𝑥)𝑑𝑥.
𝑛→∞
Using Antiderivatives to find Area
Example 1: find area on the interval  1, x . Use a geometric approach.
a) f ( x)  2
f ( x)  2
b) f ( x)  x  1
c) f ( x)  2 x  3
3
Average Value
We can also use the earlier result to find the average value of a function:
If we add up all the intermediate values of the function, and divide by the number of
intervals, we obtain:
𝑛
1
∑ 𝑓(𝑥𝑘 )
𝑛
𝑘=1
and since ∆𝑥 =
𝑏−𝑎
𝑛
, thus
∆𝑥
𝑏−𝑎
=
1
𝑛
and we have:
𝑛
1
∑(𝑓(𝑥𝑘 )∆𝑥)
𝑏−𝑎
𝑘=1
We can use this equation to find the average value of a function with n intervals.
Ex. a) Find the area under the curve of 𝑓(𝑥) = 2𝑥 2 in the interval [1,2] using 𝑛 = 3.
b) Find the average value of the function.
4
INDEFINITE INTEGRATION AS ANTI-DIFFERENTIATION
A function F is called an antiderivative of a function f on a given interval I if F ( x)  f ( x)
for all x on the interval.
Ex. 1) F ( x) 
1 3
x is an antiderivative of x 2
3
F ' ( x)  x 2  f ( x)
However, this is not the only antiderivative of f (x) - eg.
1 3 1 3
1
x , x  2, x 3  5 are
3
3
3
all antiderivatives of f ( x)  x 2 .
The process of finding an antiderivative is called integration. C is called the constant of
integration.
  x 2 dx 
 f ( x)dx  F ( x)  C
Notation:
1 3
x C
3
This is called the indefinite integral.
Power Rule: (add one to the exponent and divide by new exponent)
x n1
 x  n 1  C
For n  1 :
Ex. 2) Find
n
a)
 x dx 
b)
x
5
c)

xdx 
3
1
dx 
But: n  1 does not fit the pattern:
1
 x dx  ln x  C
 1dx  x  C
5
Properties of the Indefinite Integral
a) A constant factor can move to the outside of the integrand sign.
 cf ( x)dx  c  f ( x)dx  cF ( x)  C
b) An antiderivative of a sum is the sum of the antiderivatives.
  f ( x)  g ( x)dx   f ( x)dx   g ( x)dx  F ( x)  G( x)  C
c) An antiderivative of a difference is the difference of the antiderivatives.
  f ( x)  g ( x)dx   f ( x)dx   g ( x)dx  F ( x)  G( x)  C
Trigonometric Integration
 cos xdx  sin x  C
 sin xdx   cos x  C
Logarithmic Integration
e
x
dx  e x  C
Ex. 3) Find
 3x  x
Ex. 4) Find
 4 cos xdx 
2
 2 dx 
Integration by Substitution
For composite functions we can use substitution to make integration simpler.
Ex. 5) Find  sin( 5 x  2) dx
6
Try the following examples:
3 t
 t e dt =
4
1
sin  
 x
 3x 2 dx =

x2  1
x3  3x
dx =
Note: In general:
1
 sin( kx)dx   k cos(kx)  C
1
 cos(kx)dx  k sin( kx)  C
e
kx
dx 
1 kx
e C
k
Some useful Theorems:
Rolle’s Theorem
A function ℎ(𝑥), continuous over an interval [𝑎, 𝑏] and differentiable over ]𝑎, 𝑏[, satisfies:
If ℎ(𝑎) = ℎ(𝑏) then there must exist a point 𝑐 ∈ ]𝑎, 𝑏[, such that ℎ′ (𝑐) = 0.
What does this mean?
7
The Mean-Value Theorem
A function 𝐻(𝑥), continuous over an interval [𝑎, 𝑏] and differentiable over ]𝑎, 𝑏[, satisfies:
𝐻(𝑏) − 𝐻(𝑎) = (𝑏 − 𝑎)𝐻′(𝑐) for some 𝑐 ∈ ]𝑎, 𝑏[
or, restated:
𝐻(𝑏) − 𝐻(𝑎)
= 𝐻′(𝑐)
𝑏−𝑎
(Note: ]𝑎, 𝑏[ same as (𝑎, 𝑏))
What does this mean?
The Intermediate Value Theorem
For a function 𝑓(𝑥), continuous over an interval [𝑎, 𝑏] then for every d between 𝑓(𝑎)and
𝑓(𝑏), there exists a value c between a and b such that 𝑓(𝑐) = 𝑑.
8
Integration with a Boundary Condition and The Definite Integral
INITIAL VALUE PROBLEMS
When we integrate a function f (x) we get a family of functions F ( x )  C , but sometimes
we are interested in one particular function. We can identify this function by specifying a
specific point it passes through.
Ex. f ( x)  3  2 x  x 2
F (0)  1
This question can also be presented as
dy
 3  2x  x 2
dx
y ( 0)  1
(where the unknown is not a number, but a function)
This type of problem (involving the derivative of an unknown function) is called a
differential equation.
THE DEFINITE INTEGRAL
Area Interpretation
positive area
A1
= A1  A3  A2
x
A3
a
A2
NET Area = Area over x-axis – Area under x-axis
b
negative area
9
𝑏
Notation: Recall: lim ∑𝑛𝑘=1(𝑓(𝑥𝑘 )∆𝑥) which is approximately equal to: ∫𝑎 𝑓(𝑥)𝑑𝑥.
𝑛→∞
b
 f ( x)dx
a
is the definite integral of f from a to b where: a is the lower limit, b is the upper limit, f (x) is
the integrand.
Ex. Sketch and evaluate using geometric formulae.
4
 2dx
1)
1
2
 ( x  2)dx
2)
1
2
 ( x  1)dx
3)
Note: Must split into two integrals. (Net area is 0).
0
A( x)  f ( x)
b
A(a )  0
A
A   f ( x)dx
a
A(b)  A
a
b
A( x)  f ( x)
let F ( x)  A( x)  C
Area = F (b)  F (a)  A(b)  c  A(a)  c  A(b)  A(a)  A  0  A
b
So,
 f ( x)dx  F (b)  F (a)
a
10
The Fundamental Theorem of Calculus Part 1:
If f if continuous on a, b and F is the antiderivative of f on a, b , then:
b
 f ( x)dx  F (b)  F (a) = F ( x)
b
a
or F ( x)a
b
a
Refer to old examples.
Basic Properties:
a
 f ( x)dx  0
a
b

a
f ( x)dx    f ( x)dx
a
b
b
b
a
a
 cf ( x)dx  c f ( x)dx
b
b
b
a
a
a
  f ( x)  g ( x)dx   f ( x)dx   g ( x)dx
b

c
b
a
c
f ( x)dx   f ( x)dx   f ( x)dx
a
acb
Examples:
1
1)
x
2
dx
1
2
2)
e
3x
dx
1

3)
4
 3 cos( 2 x)dx
0
3
4)
 (x  e
2 x
) dx
1
11
m
5) Find value(s) of m where  (2 x  4)dx  5 .
0
6) Find the total area of the region enclose by y  x 3  1 , the x-axis, and 0  x  2 . (Note:
need to consider net signed areas). (Split into two integrals and take the negative of the
negative area).
7). a) Find the area under the curve of 𝑓(𝑥) = 2𝑥 2 in the interval [1,2].
b) Find the average value of the function.
12
More Integration by Substitution
PART I – INDEFINITE INTEGRALS
EXAMPLES : Find the indefinite integral of the following using the method of substitution:
1)
 2x  x
2
 9  dx
5
2)
4x
 5  3x
2
dx
PART II – DEFINITE INTEGRALS
EXAMPLES : Evaluate the following definite integrals using the method of substitution

1
1)
2
 2 x 1  x dx
0
2
2)
cos x
 1  2sin x dx
0
13
4e
3)
1
 x  ln x 
e
3
dx
4) Find

1  x 2 dx using the substitution x  sin  .
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FURTHER INTEGRATION TECHNIQUES – INTEGRATION BY PARTS
OBJECTIVE  Use further integration techniques including integration by substitution, integration by parts
and repeated integration by parts to integrate more difficult integrands
PART I – DERIVING THE FORMULA
Consider the product of two functions,
uv .
1) Differentiate the product with respect to x.
(uv)  u v  uv
(uv)  vdu  udv
2) Integrate both sides of the equation you found in 1).
 (uv)   (vdu  udv)
uv   vdu   udv
3) Rearrange the equation in 2) to obtain the “PARTS FORMULA” :
uv   vdu   udv
uv   vdu   udv
SUMMARY :
 udv  uv   vdu
PART II – EXAMPLES
Find the following integrals.
1)
 x cos xdx
2)
 xe dx
x
15
3)
x
4)
3e
5)
 x cos 2 dx
6)
e
3
x
ln xdx
2x
dx
x
2x
x
sin dx
3
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REVISITING KINEMATICS AND AREAS BETWEEN CURVES
Kinematics
Recall that the gradient (slope) of a displacement-time graph gives a velocity-time graph.
Therefore, the area under a velocity-time graph gives a displacement-time graph and the area
under an acceleration-time graph gives a velocity time graph.
Using this process you can derive the formulas for velocity and displacement given uniformly
accelerated motion (physics).
Note: Displacement = total area
Distance = net signed area
a
b
Ex. The velocity of an object is v m/s after t seconds, where v  1 2 sin 2t .
a) Find the object’s displacement at

seconds.
4
b) Find the object’s distance over the first

seconds. (Absolute Value)
4
(Determine t-intercepts)
17
Areas Between Two Curves
We can find the area between a curve and the x-axis, now we will find the area between two
curves.
b
A    f ( x)  g ( x)dx
a
b
b
a
a
A   f ( x)dx   g ( x)dx
a
b
Ex. Find the area of the region bounded by the parabolas y  x 2 and y  2 x  x 2 .
First find the intersection (endpoints).
Look at the graph to determine which function is f (x) and which is g (x ) .
18
Ex. Find the area of the region enclosed by x  y 2 and y  x  2 .
Draw a sketch because we have two separate regions.
A1
A2
(Note: 2 possible ways: Find endpoints in terms of x and integrate dx, or find endpoints in
terms of y and integrate dy.)
19
VOLUMES OF REVOLUTION
If a curve with fixed boundaries is rotated around the x-axis, a 3-dimensional solid is formed.
For example:
Take the function y  x , 0  x  1 .
If the curve is rotated around the xaxis, a cone is formed. We can find
the volume of this shape by slicing it
into cylindrical sections (circles) and
summing their areas.
yx
The volume is the product of the area times the height.
V  A  h  (area of a disk)x(height)
Each circle has radius y as x goes from zero to 1.
The area of one circle (disk) is y 2 . The width of each circle is the change in x from one
circle to another.
Therefore, we can sum the areas of the circles using the formula:
n
V   A( x)  x ,
where n is the number of circles and x is the width of each
k 1
circle.
If we allow the number of intervals to increase, the width of the intervals to approach
zero, then the limit of this sum as x  0 is:
b
b
V   A( x)dx     f ( x) dx
2
a
a
20
For each disk, the area is y 2 , but using the function, we know that y  x , therefore we can
 x3
write: V   x dx   
 3
0

1
2

    1  0    .

3
 3
0
1
Notice that we got the volume of the shape to be
1 2
r h , which happens to be the equation
3
for volume of a cone.
When a plane region enclosed by the curve
about the x-axis, the volume, V
y  f  x
and the lines
and
x b
is revolved
and
x b
is revolved
units 3 , of the solid formed is given by :
V 
x b
b
  f ( x)  dx
2
or
xa
When a plane region enclosed by the curve
about the y-axis, the volume, V
xa
units
V 
y f

3
V    y 2 dx
a
y  f  x
and the lines
xa
, of the solid formed is given by :
 f 1 ( y)  dy
f
2
or
y e
V    x 2 dy
e
A solid of revolution is one generated by rotating a plane region a out a line that lies in the
same plane as the region. For example, if the line y  3 , 2  x  4 , is rotated about the xaxis, what solid is formed?
y3
2
4
4
4
4
4
 
V   A( x)dx   y 2 dx    3 2 dx    9dx  9 x 2  9 (4  2)  9  2  18
4
2
2
2
2
21
Ex. Write the integral for the volume of the plane region defined by the function y  r from
x=a to x=b
Ex. Find the volume of the solid obtained by rotating about the x-axis the region under the
curve y  x from 0 to 1.
Ex. The curve y  x  1,1  x  5 is rotated about the x-axis to form a solid of
revolution. Sketch this solid and find its volume.
Ex. The curve y  x  1,1  x  5 is rotated about the y-axis. Sketch this second solid
and find its volume.
22
Ex. Find the volume of the solid of revolution formed by rotating the part of the curve
y  e x between x=1 and x=5 about the y-axis.
Ex. Find the volume of the solid formed by revolving the region enclosed by the curve
f  x   25  x2 and the line g  x   3 about the x-axis.
23