Lesson 2
The Definite Integral and Area
Last Time…
• Approximate total change from rate of change using
(LHS/RHS)
• For increasing functions, left hand sums underestimate
and right hand overestimate.
• LHS: multiply function at left of each subinterval by
width of subinterval.
• Riemann sums look like areas of rectangles.
Example: Velocity and Distance Traveled
Riemann sum rectangles, ∆t = 4 and n = 1:
Riemann sum rectangles, ∆t = 2 and n = 2:
Riemann sum rectangles, n = 5 (∆t varies)
Review
1- The main question answered in §5.1 is:
rate of
If we know the __________
change, how can we
total
estimate the __________
change?
2- If we use a smaller time interval, the upper
(a)
and lower estimates get _________
(a) closer together
(b) farther apart
§5.(2-3): The Definite Integral and Area
Estimating Total Change
Let f(t) be a rate of change on [a,b]:
Split [a, b]
into subintervals:
Add up:
LHS:
RHS:
"t =
b!a
n
Getting Better Estimates
• Average the left and right hand sums
• Use more subintervals!
The Definite Integral
Let f(t) be continuous. As n increases,
the approximations get better. Define:
The definite integral of f(t) from t = a to
t = b.
Example:
If v(t) is velocity (ft/sec) at time t (sec),
then:
Represents the total
change in position from
time t = 1 to t = 3
seconds.
Computing Integrals
•
•
•
•
Can’t compute exactly–yet!
Left hand sum
Right hand sum
Average
Example:
t
0
0.5
1
1.5
2
g(t)
1
2
1
0
2
Estimate:
Left Hand Sum: 0.5(1 + 2 + 1 + 0) = 2
Right Hand Sum: 0.5(2 + 1 + 0 + 2) = 2.5
Average: (2 + 2.5)/2 = 2.25 Best Estimate
so far!
Example:
Pictures:
t
0
0.5
1
1.5
2
g(t)
1
2
1
0
2
Example:
Estimate, using n = 4 subintervals:
0.5 .
Here, ∆t = ______
In fact ∆t = (2-0)/4 = 0.5 .
Use a left hand sum or a right hand sum…
LHS: 3.75.
RHS: 5.75.
Average: 4.75
(Actual: 4.667)
Example:
Pictures, using n = 4:
Left Hand Sum:
Right Hand Sum:
The Integral as Area
Let f(x) ≥ 0.
As n increases…
The Integral as Area
Let f(x) ≥ 0.
As n increases…
the rectangles
approach the area
under the curve!
The Integral as Area
If f(x) ≥ 0, then
represents the area underneath the curve
f between x = a and x = b.
We can use this to calculate integrals!
Example:
Find:
Start with a sketch.
Count boxes: 2.5
Area of each box: 1
Area under curve: 2.5
Example:
Estimate:
Note: you have to deal
with “partial” boxes.
I estimate about 4
boxes.
Area of each box? 1
So Area =
~4
Area Below the Axis
If the function is sometimes negative:
How do we find total area?
How do we find total change (the definite integral)?
Area Below the Axis
For a general function:
(Positive)
Total
Change:
-
(Positive)
NOTE:
Total Area=
A1 + A2
Example:
Positive or negative?
Positive, because
most area is above
the x-axis.
Summary
• Defined the definite integral of a continuous
function (limit of Riemann sums)
• Integral of a rate of change is the total
change
• Estimated integrals with Riemann sums
• Interpreted integrals as areas, including
“negative area”
• Evaluated integrals with area arguments
Practice
~ 5.7
~9.5
a. Estimate (by counting the squares) the total area between
f(x) and the x-axis.
8
b. Using the given graph, estimate ! 0 f (x)dx
c. Why are your answers in parts (a) and (b) different?
Practice
~ 5.7
~9.5
a. The total area between f(x) and the x-axis:
~ 5.7 + 9.5=15.2
b. Using the given graph, estimate ! 0 f (x)dx ~5.7 c. Why are your answers in parts (a) and (b) different?
Total area is different from Total change
8
9.5 = -3.8
Group work
• Find the area under the graph of y =x2 on the
interval [1, 3] with n = 2 using left
rectangles.
• Is this estimate an under or over estimate?
(Hint: Consider the graph of the function
with the rectangles.)
• Repeat the estimate with right rectangles.
• Find the average of the two estimates.
Group work
• Find the area under the graph of y =x2 on the
interval [1, 3] with n = 2 using left rectangles.
A = 1*(1+4) = 5
Is this estimate an under or over estimate? (Hint:
Consider the graph of the function with the
rectangles.)
This is an underestimate
• Repeat the estimate with right rectangles.
A = 1*(4+9) = 13
• Find the average of the two estimates.
(5+13)/2 = 9
L
R