F1 – Definite Integrals - Area
under Curves as Limits and Sums
IB Mathematics HL/SL & MCB4U
(A) Review
• We have looked at the process of anti-differentiation (given
the derivative, can we find the “original” equation?)
• Then we introduced the indefinite integral  which
basically involved the same concept of finding an
“original”equation since we could view the given equation
as a derivative
• We introduced the integration symbol  
• Now we will move onto a second type of integral  the
definite integral
(B) The Area Problem
• to introduce the second kind of integral : Definite
Integrals  we will take a look at “the Area
Problem”  the area problem is to definite
integrals what the tangent and rate of change
problems are to derivatives.
• The area problem will give us one of the
interpretations of a definite integral and it will lead
us to the definition of the definite integral.
(B) The Area Problem
• Let’s work with a
simple quadratic
function, f(x) = x2 + 2
and use a specific
interval of [0,3]
• Now we wish to find
the area under this
curve
(C) The Area Problem – An
Example
• To estimate the area under the curve, we will divide the are into simple
rectangles as we can easily find the area of rectangles  A = l × w
• Each rectangle will have a width of x which we calculate as (b – a)/n
where b represents the higher bound on the area (i.e. x = 3) and a
represents the lower bound on the area (i.e. x = 0) and n represents the
number of rectangles we want to construct
• The height of each rectangle is then simply calculated using the
function equation
• Then the total area (as an estimate) is determined as we sum the areas
of the numerous rectangles we have created under the curve
• AT = A1 + A2 + A3 + ….. + An
• We can visualize the process on the next slide
(C) The Area Problem – An
Example
• We have chosen to draw 6
rectangles on the interval [0,3]
• A1 = ½ × f(½) = 1.125
• A2 = ½ × f(1) = 1.5
• A3 = ½ × f(1½) = 2.125
• A4 = ½ × f(2) = 3
• A5 = ½ × f(2½) = 4.125
• A6 = ½ × f(3) = 5.5
• AT = 17.375 square units
• So our estimate is 17.375 which
is obviously an overestimate
(C) The Area Problem – An
Example
• In our previous slide, we used 6 rectangles which
were constructed using a “right end point” (realize
that both the use of 6 rectangles and the right end
point are arbitrary!)  in an increasing function
like f(x) = x2 + 2 this creates an over-estimate of
the area under the curve
• So let’s change from the right end point to the left
end point and see what happens
(C) The Area Problem – An
Example
• We have chosen to draw 6
rectangles on the interval [0,3]
• A1 = ½ × f(0) = 1
• A2 = ½ × f(½) = 1.125
• A3 = ½ × f(1) = 1.5
• A4 = ½ × f(1½) = 2.125
• A5 = ½ × f(2) = 3
• A6 = ½ × f(2½) = 4.125
• AT = 12.875 square units
• So our estimate is 12.875 which
is obviously an under-estimate
(C) The Area Problem – An
Example
• So our “left end point” method (now called a left hand
Riemann sum) gives us an underestimate (in this example)
• Our “right end point” method (now called a right handed
Riemann sum) gives us an overestimate (in this example)
• We can adjust our strategy in a variety of ways  one is
by adjusting the “end point”  why not simply use a
“midpoint” in each interval and get a mix of over- and
under-estimates?  see next slide
(C) The Area Problem – An
Example
• We have chosen to draw 6
rectangles on the interval [0,3]
• A1 = ½ × f(¼) = 1.03125
• A2 = ½ × f (¾) = 1.28125
• A3 = ½ × f(1¼) = 1.78125
• A4 = ½ × f(1¾) = 2.53125
• A5 = ½ × f(2¼) = 3.53125
• A6 = ½ × f(2¾) = 4.78125
• AT = 14.9375 square units
which is a more accurate
estimate (15 is the exact
answer)
(C) The Area Problem – An Example
•
We have chosen to draw 6
trapezoids on the interval [0,3]
•
•
•
•
•
•
•
A1 = ½ × ½[f(0) + f(½)] = 1.0625
A2 = ½ × ½[f(½) + f(1)] = 1.3125
A3 = ½ × ½[f(1) + f(1½)] = 1.8125
A4 = ½ × ½[f(1½) + f(2)] = 2.5625
A5 = ½ × ½[f(0) + f(½)] = 3.5625
A6 = ½ × ½[f(0) + f(½)] = 4.8125
AT = 15.125 square units
•
(15 is the exact answer)
(D) The Area Problem – Expanding
our Example
• Now back to our left and right Riemann
sums and our original example  how can
we increase the accuracy of our estimate?
• We simply increase the number of
rectangles that we construct under the curve
• Initially we chose 6, now let’s choose a few
more … say 12, 60, and 300 ….
(D) The Area Problem – Expanding
our Example
# of rectangles
Area estimate
12
16.15625
60
15.22625
300
15.04505
(D) The Area Problem – Expanding
our Example
# of rectangles
Area estimate
12
13.90625
60
14.77625
300
14.95505
(E) The Area Problem - Conclusion
• We have seen the following general formula used in
the preceding examples:
• A = f(x1)x + f(x2)x + …. + f(xi)x + ….. + f(xn)x
as we have created n rectangles
• Since this represents a sum, we can use summation
n
notation to re-express this formula  A 
f ( x )x

i 1
• So this is the formula for our Riemann sum
i
(F) Riemann Sums – Internet
Interactive Example
• Visual Calculus - Riemann Sums
• And some further worked examples
showing both a graphic and algebraic
representation:
• Visual Calculus - Riemann Sums – 2
(G) The Area Problem – Further
Examples
• (i) Find the area between the curve f(x) = x3 – 5x2 + 6x + 5
and the x-axis on [0,4] using 5 intervals and using rightand left- and midpoint Riemann sums. Verify with
technology.
• (ii) Find the area between the curve f(x) = x2 – 4 and the
x-axis on [0,2] using 8 intervals and using right- and leftand midpoint Riemann sums. Verify with technology.
• (iii) Find the area between the curve f(x) = x2 – 2 and the
x-axis on [0,2] using 8 intervals and using midpoint
Riemann sums. Verify with technology.
(G) The Area Problem – Further
Examples
• Homework to reinforce these concepts:
• Stewart, 1989, Chap 10.4, p474, Q3c,4d,5
• Stewart, 1989, Chap 10.5, p482, Q1,3
(H) The Area Problem – Exact Areas
• Now to make our estimate more accurate, we simply made more
rectangles  how many more though?  why not an infinite amount
(use the limit concept as we did with our tangent/secant issue in
differential calculus!)
• Then we arrive at the following “formula”:
A  lim
n 
n
 f ( x )x
i 1
i
(H) The Area Problem – Areas as
Limits
• The formula we will
n
use is
A  lim
n 
 f ( x )x
i 1
i
• The function will be
f(x) = x3 – x2 - 2x - 1
on [0,2]
• Which we can graph
and see 
(H) The Area Problem – Areas as Limits
• So our formula is:
A  lim
n 
n
 f ( x )x
i 1
i
• Now we need to work out just what f(xi) and x are equal to
so we can sub them into our formula:
• x = (b-a)/n = (2-0)/n = 2/n
• xi simply refers to the any endpoint on any one of the many
rectangles  so let’s work with the ith endpoint on the ith
rectangle  so in general, xi = a + ix
(H) The Area Problem – Areas as Limits
• I am using a + idx on
the diagram to
represent the ith
rectangle, which is ix
(or idx) units away
from a
• Then the height of this
rectangle is f(a + ix)
(H) The Area Problem – Areas as Limits
•
Now back to the formula in which x = 2/n and xi = 0 + i x or xi = 2i/n
A  lim
n 
A  lim
n 
n
 f ( x )x
i
i 1
 x
n
i 1
i
3

 2
2
 xi  2 xi  1  
n
  2i 3  2i  2  2i   2 
A  lim         2   1 

n 
n
 n   n 
i 1   n 
n 
 16i 3   8i 2   8i1   2  
A  lim    4    3    2     
n 
i 1   n
  n   n   n 
 16 n 3 8 n 2 8 n 1 2 4 
A  lim  4  i  3  i  2  i  1
n  n
n i 1
n i 1
n i 1 
 i 1
n
(H) The Area Problem – Sums of Powers
• We have the following summation formulas to use in our
simplification:

n
1 2
i  n n

2
i 1
n





1
3
2
i  2n  3n  n

6
i 1
2
n
1 4
3
2
i

n

2
n

n

4
i 1
3
n

1
5
4
3
i

6
n

15
n

10
n
n

30
i 1
4

(H) The Area Problem – Areas as Limits
• Now we simply substitute our appropriate power sums:
A  lim
n 
n
 f ( x )x
i 1
i
16 n 3 8 n 2 8 n 1 2 4 
A  lim  4  i  3  i  2  i  1
n  n
n i 1
n i 1
n i 1 
 i 1
16 n 4  2n 3  n 2 8 2n 3  3n 2  n 8 n 2  n 2 n 
A  lim  4 
 3
 2
  
n  n
4
n
6
n
2
n 1

8
8 8 2
16 32 16 16 24
A  lim  
 2 
 2 
 
n  4
4n 4n
6 6n 6n 2 2n 1 

8 
 16

 32 24 8 
 16
A  lim 4   4  2  lim  
   lim  2  2 
n 
6

 n   4n 6n 2n  n   n 6n 
2
A  4
3
(H) The Area Problem – Areas as Limits
• Now we can confirm this
visually using graphing
software (WINPLOT):
• And we get the total area
between the axis and the
curve to be -4.666666666
from the software!
(H) The Area Problem – Areas as Limits
– Additional Examples
• (i) Find the area between the x-axis and the
function f(x) = x2 on [0,2] using the areas as
limits approach. Confirm with graphing
technology
• (ii) Find the area between the x-axis and the
function f(x) = x3 + x on [1,4] using the
areas as limits approach. Confirm with
graphing technology
(I) Areas as Limits - Homework
• Homework to reinforce these concepts:
• Stewart, 1989, Chap 10.4, p474, Q3,4,6*
Internet Links
• Calculus I (Math 2413) - Integrals - Area
Problem from Paul Dawkins
• Integration Concepts from Visual Calculus
• Areas and Riemann Sums from P.K. Ving Calculus I - Problems and Solutions