Math 2015
Lesson 2
The Definite Integral and Area
Last class, we saw that we could estimate the total change in a quantity from its rate of
change using a left or right hand sum. This fact will motivate us in defining the so-called
definite integral, which will be our primary topic of study in this course.
Estimating Total Change From Rates of Change
Let’s review what we did last class. Let f(t) be a rate of change on a ≤ t ≤ b.
To approximate the total change from a to b:
1. Split [a,b] into n subintervals, each of width ∆t = (b – a)/n. (See diagram below.)
Let the points subdividing the intervals be labeled t0, t1, ... , tn, with t0 = a and
tn = b. So then the first interval is [t0, t1], the second is [t1, t2], and the last is [tn–
1, tn].
2. Approximate the change on the ith subinterval [ti-1, ti] with either
o f at the left side times ∆t: f(ti-1) ∆t, which makes a left hand sum; or
o f at the right side times ∆t: f(ti) ∆t, which makes a right hand sum.
3. We approximate the total change by adding up the estimated change on each
subinterval.
So we get either

A Left Hand Sum:
(The sigma just means “add up everything like this starting at i = 0 and ending when
you get to i = n–1.” You will not need to worry much about sigma notation; it’s just
shorthand which means to add things up.)

or a Right Hand Sum:
We know we can probably get an even better approximation by averaging the two. We
also know we can get better approximations by using more subintervals, which means
using a larger n, and a smaller ∆t.
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Math 2015
Lesson 2
The Definite Integral
Suppose we have a continuous function f. (Think, “nice function.”) Let n get very large,
so that ∆t = (b – a)/n becomes very small. The idea is that as n gets huge, our left and
right hand sums should be very close to the actual change on [a, b]. Therefore, we define
the definite integral of f from a to b as follows:

b
a
f (t)dt  lim (LHS)  lim (RHS)
n
n
where LHS is the Left Hand Sum and RHS is the Right Hand Sum with n subintervals.
In this notation, the integral sign is an elongated “S”, which of course stands for sum.
(Really!) The a and b are the lower and upper limits, and tell us what interval we were
working over. The little dt at the end is called a differential. (You studied differentials in
a different context in 1016.) The differential marks the end of the integral, and tells us
that the variable we used is t; it also takes the place of the ∆t in the sum. (We will find
more use for the dt later.)
Thus, the definite integral should be exactly the total change, and this is the quantity we
are approximating using left and right hand sums.
If v(t) is the velocity (in ft/sec) at time t (in seconds), then 1 v(t) dt is the
total change in position from t = 1 to t = 3. (Remember, velocity is the
rate of change of position with respect to time, so the integral gives total
change in position.)
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Example:
Computing Integrals
We don’t yet know a way to compute integrals exactly, but we can estimate them using
left and right hand sums, or the average of both.
Example:
Suppose we have data for a function g(t) as given below:
t
0 0.5 1 1.5 2
g(t) 1 2
1 0
2
Estimate 0 g(t) dt . What is ∆t here? ∆t = _____.
Using a left hand sum:
2
(Note: Since each function value is multiplied by ∆t, and since ∆t = 0.5 for all
subintervals, we can multiply after adding up the left hand side of each interval.)
Using a right hand sum:
Or the average:
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
Math 2015
Lesson 2
The pictures are shown below. Note that in each, one of the “rectangles”
has no height, because one of the function values is 0!
Are either of these over or underestimates? Who knows—the function is
neither always increasing nor always decreasing.
Example:
Estimate
 t
2
0
2
1dt with n = 4 subintervals.
Here, ∆t = _______________.

Left Hand Sum:
Right Hand Sum:
In this case, the average of the left and right hand sums is 4.75. It turns
out that the actual value of the definite integral is 4 and 2/3, or about
4.6667, so we have a pretty good estimate.
Sketch in the pictures for the left and right hand sums below:
Left Hand Sum:
Right Hand Sum:
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Math 2015
Lesson 2
The Integral as Area Under a Curve
Suppose we have a curve f(x) which is always positive. We imagine what happens as we
set up rectangles for right hand sums using increasing values of n, as shown below:
n=5
n = 10
n = 20
(We could have used left hand sums instead, of course.) As the number of rectangles goes
to infinity and they become very skinny, we see that the total area encompassed by the
rectangles comes close to the area underneath the curve. Since Riemann sums are really
the areas of the rectangles, we have determined that:
If f(x) ≥ 0, then

b
a
f (x)dx represents the area under the curve f between
x = a and x = b.
Evaluating Integrals using Area
Since we know that for positive functions, the integral is just the area under the curve, we
can use basic geometry to calculate some integrals.
Example:
Calculate

3
0
x  1 dx .
We begin with a sketch of the
curve. (Note that x – 1 would be a
straight line with slope 1 through
the point x = 1 on the axis. The
absolute values simply “flip up” the
negative values to become positive.)
We have plotted this at right. To
help, we have included grid lines which show where different x and y
values occur.
Using this sketch, we estimate the total area under the curve to be
3
________ units. Therefore, 0 x  1 dx 
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Math 2015
Example:
Lesson 2
3
Estimate 0 x dx by counting boxes. (Note: You will have to decide
how to count “partial” boxes.
2
Estimate:
(Note that the x and y axes have
different scales!)
(You could also estimate this by
using left or right hand sums.)
Areas Below the x-Axis
What if f is sometimes negative? Then some elements of the left or right hand sums will
have a negative value times a width, and part of the sum may be negative. (This has the
effect of saying that rectangles below the x-axis have negative area, which of course is
impossible. Nonetheless, we will sometimes refer to “negative areas” below the x-axis as
a sort of shorthand description of the situation.)
Therefore, if a function has both negative and positive values between x = a and x = b,
then we say

b
a
f (x)dx  (Area above the axis) – (Area below the axis)
In all cases, we are referring to the area between the curve and the x-axis. (Think of
where the rectangles in a left or right hand sum would be.)
For example, in the picture below, if A1 is the area under the first “hump” (above the xb
axis) and A2 is the area under the second (below the x-axis), then a f (x)dx  A1  A2 .
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Math 2015
Example:
Lesson 2
Determine whether
 5  e dx is positive or negative.
2
x
0
The graph of
y  5  e x is shown below.


The integral must be ____________, since most of the area between the
curve and the axis is ___________ the x-axis.
Summary
Today, we have

Defined the definite integral of a continuous function in terms of the limit of left
and right hand sums.

Given an interpretation to a definite integral. Namely, if f is a rate of change of
b
some quantity, then a f (t)dt is the total change in that quantity on the interval
t = a to t = b.

Estimated definite integrals using left and right hand sums. Given an
interpretation of the definite integral as area underneath a curve if the integrand is
positive

Determined that area beneath the x-axis is treated as “negative area.”

Used area arguments to evaluate definite integrals.
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