AP Calculus AB
5.2 Definite Integrals
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Name__________________________________
I can express the area under a curve as a definite integral and as a limit of Riemann sums
I can compute the area under a curve using numerical integration procedures
In the previous section, we estimated distances and areas with finite sums, using LRAM, RRAM, and
MRAM methods. In this section, we move beyond finite sums to see what happens in the limit as the terms
become infinitely small and their numbers infinitely large.
In order to compute these sums, we will need to use Signma-notation for a sum, so let’s quickly review
Signma-notation:
n
a
k 1
k
 a1  a 2  a3  . . .  a n 1  a n
Riemann Sums
Georg Friedrich Bernhard Riemann (nothing like having a four-part
name) lived from 1826-1866. The type of sum we are going to study are
named after him because, while the limits of the sums we are about to
learn about were used well before Riemann’s time, they were used
without mathematical proof until Riemann was able to prove the
existence of their limit in 1854. The proof of the existence of Riemann
sums is beyond the scope of this course (that is one of my favorite
phrases, because it means that we do not have to worry about the proof).
In the previous section, we used rectangles to approximate area under a curve. While all RAM sums are
technically Riemann sums due to the way they are constructed, there are ways to construct Riemann sums so
that the sums may be used for all types of functions and not just nonnegative functions.
The process to find a Riemann Sum is described below.
Look at y  f  x  on the closed interval  a, b
 Divide  a, b into n subintervals by choosing n 1
points, say x1 , x2 ,..., xn 1 .
 The intervals DO NOT need to be the same length.
To keep notation consistent, let a  x0 and b  xn
 So the width of the 3rd interval is x3  x2 ; similarly,
the length of the kth interval is xk  xk 1
 For the kth subinterval, its width xk is given by
xk  xk  xk 1
 In each subinterval xk , pick some point ck . It
DOES NOT matter where in the interval you pick ck
 On each subinterval, stand a rectangle that reaches
from the x-axis to touch the curve at  ck , f  ck   ;
These rectangles can lie either above or below the xaxis
 The area of each rectangle is f  ck   xk and you
will have n total rectangles
 The sum of the area of all the rectangles will be
n
Sn   f  ck   xk
k 1
The sum S n , which depends on the partition P and the choice of numbers ck , is a
Riemann Sum for f on the interval [a, b]
As the partitions  a, b become finer and finer, we would expect the rectangles defined by the partitions
to approximate the region between the x-axis and the graph of f with increasing accuracy.
The largest subinterval length of the partition P (remember, each partition can be a different length) is
called the norm of the partition, and is denoted P . As P  0 (or alternately as n   ), all Riemann
Sums for a given function will converge to a common value.
The Definite Integral as a Limit of Riemann Sums
Let f be a function defined on a closed interval  a, b . For any partition P of  a, b , let the numbers ck
be chosen arbitrarily in the subintervals  xk 1 , xk  . If there exists a number I such that
n
lim  f  ck   xk  I
P 0
k 1
no matter how P and the ck ' s are chosen, then f is integrable on  a, b and I is the definite integral of f
over  a, b .
Theorem: All continuous functions are integrable. That is, if f is continuous on  a, b , then its definite
integral over  a, b exists
The Definite Integral of a Continuous Function on  a, b
Let f be continuous on  a, b and let  a, b be partitioned into n subintervals of equal length
x   b  a  / n . Then the definite integral of f over  a, b is given by
n
lim  f  ck   xk
n 
th
k 1
Where each ck is chosen arbitrarily in the k subinterval.
The definite integral is the “signed” area under a curve f ( x) from a to b
Integral Notation
Example 1
Evaluate
Example 2
0  x  2 dx
3
Example 3
Evaluate

3
1
5 dx
Evaluate

2
2
4  x 2 dx
More About Signed Area
Note that all the functions we have talked about so far have been nonnegative. However, we stated that the
integral is the signed area under a curve. Therefore, if a function f  x   0 , then its area will be negative.
Therefore,
 f  x  dx    area  if f  x   0
b
a
or more commonly
Area   f  x  dx if f  x   0
b
a
If an integrable function y  f  x  has both positive and negative values on an interval  a, b , then
 f  x  dx 
b
a
(area above the x-axis)  (area below the x-axis)
Exploration
It is a fact that


0
sin x dx  2 .
With that information, determine the values of the following integrals.
2
1.

3.

 /2
5.



2
7.

sin x dx
4.
  2  sin x  dx
2sin u du
6.

2


0
0
0
2
2.
sin x dx
x
sin   dx
2
8.
0
sin x dx

0
2
0
sin  x  2 dx
cost dt
Integrals on a Calculator
Not surprisingly, your calculator is able to handle numeric integrals. To calculate numeric integrals on your
TI-Nspire CX, simply use the
button and find the integral template. It looks like this:
Alternatively, you can also click menu 4:Calculus 2:Numerical Integral to pull up the template.
Practice
Evaluate the following integrals on your calculator:
1.

2
1
x sin x dx
2.
1
4
 1 x
0
2
dx
3.

5
0
et dt
2
Functions with Discontinuities
Earlier, we had a theorem that stated that all continuous functions are integrable. But what about
discontinuous functions? As it turns out, some of them are integrable also.
Example 3
Find
2
x
1
x

dx
Exploration 2
1.
x2  4
Explain why the function f  x  
is not continuous on the interval 0,3 . What kind of
x2
discontinuity occurs?
2.
Show that
3.
Show that

3

5
0
0
x2  4
dx  10.5
x2
int( x) dx  10