Download Calculus 7.1A lesson notes

Chapter 7
Additional
Integration
Topics
Section 1
Area Between Curves
Learning Objectives for Section 7.1
Area Between Two Curves
The student will be able to:
Day 1:
•
Determine the unsigned
area between a curve and
the x axis.
• Determine the unsigned
area between two curves.
Day 2:
• Solve applications involving
area between curves
Barnett/Ziegler/Byleen Business Calculus 12e
2
Area “Between Curves”
 In the previous chapter, we found that the definite integral
𝑏
𝑓 𝑥 𝑑𝑥
𝑎
represents the accumulated sum of the signed areas between
the graph of f(x) and the x-axis over the interval [a, b].
 In this chapter:
• We will treat all areas as positive, even though they
may fall below the x-axis.
• We will find the total unsigned area between two
curves.
Barnett/Ziegler/Byleen Business Calculus 12e
3
No Negative Areas!
The unsigned area between the
graph of a negative function and the
x axis is equal to the definite
integral of the negative of the
function.
A

b
a
[  f ( x) ] dx
B
𝐴=−

c
b
B
a
A
𝑏
𝑂𝑅
y = f (x)
b
c
𝑓 𝑥 𝑑𝑥
𝑎
f ( x) dx
Barnett/Ziegler/Byleen Business Calculus 12e
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Area Between f(x) and x-axis
The area between f (x) and the x axis can be found as follows:
If f (x) is above the x axis:
Area =

b
a
f ( x) dx
If f (x) is below the x axis:
Area =

b
a
[  f ( x) ] dx
𝑏
𝑂𝑅
𝐴𝑟𝑒𝑎 = −
𝑓 𝑥 𝑑𝑥
𝑎
Barnett/Ziegler/Byleen Business Calculus 12e
5
Example 1
Find the area bounded by
y = 4 – x2
x-axis
Find x-intercepts:
1x4
-2, 2
2
𝐴𝑟𝑒𝑎 =
4
𝑓 𝑥 𝑑𝑥 −
1
2
𝑓 𝑥 𝑑𝑥
2
4
4 − 𝑥 2 𝑑𝑥 −
=
1
𝑥3 2
= 4𝑥 −
−
3 1
=
4 − 𝑥 2 𝑑𝑥
2
𝑥3
4𝑥 −
3
8
1
8−
− 4−
3
3
−
4
2
64
8
16 −
− 8−
3
3
Barnett/Ziegler/Byleen Business Calculus 12e
37
=
3
6
Example 2
Find the area bounded by
y = x2 – 9
Find x-intercepts:
x-axis
-3, 3
Over the interval: [2, 4]
3
𝐴𝑟𝑒𝑎 = −
4
𝑓 𝑥 𝑑𝑥 +
2
3
3
4
𝑥 2 − 9 𝑑𝑥 +
=−
3
2
𝑥 2 − 9 𝑑𝑥
3
4
−𝑥 2 + 9 𝑑𝑥 +
=
2
𝑥3
3
= − + 9𝑥 +
3
2
=
𝑓 𝑥 𝑑𝑥
𝑥 2 − 9 𝑑𝑥
3
3
𝑥
4
− 9𝑥
3
3
8
−9 + 27 − − + 18 +
3
Barnett/Ziegler/Byleen Business Calculus 12e
64
− 36 − 9 − 27
3
=6
7
Area Between Two Curves
Theorem 1. If f (x)  g(x) over the
interval [a, b], then the area bounded
by y = f (x) and y = g(x) for a  x  b
is given by
A
  f ( x)  g ( x) dx
b
f (x)
A
a
b g (x)
a
Note: The graphs do not need to be above the xaxis; however, the graph of f must be above (or
equal) to the graph of g throughout the interval
[a, b].
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 3
Find the area bounded by
f(x) = x2 – 1 and
g(x) = 3.
Interval isn’t given so you need to find
where the two graphs intersect. (use
“CALC Intersect” function on calculator or
do it algebraically) (-2, 3) & (2, 3)
2
𝐴𝑟𝑒𝑎 =
2
3 − 𝑥 2 − 1 𝑑𝑥
𝑔 𝑥 − 𝑓 𝑥 𝑑𝑥 =
−2
−2
2
4 − 𝑥 2 𝑑𝑥
=
−2
3
𝑥 2
8
8
= 4𝑥 −
= 8−
− −8 +
3 −2
3
3
Barnett/Ziegler/Byleen Business Calculus 12e
32
=
3
9
Example 4
 Find the area bounded by
𝑓 𝑥 = 5 − 𝑥 2 and 𝑔 𝑥 = 2 − 2𝑥
Finding intersection points algebraically:
5 − 𝑥 2 = 2 − 2𝑥
𝑥 2 − 2𝑥 − 3 = 0
(𝑥 − 3)(𝑥 + 1) = 0
𝑥 = −1, 3
3
𝑓 𝑥 − 𝑔(𝑥) 𝑑𝑥
−1
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 4 (continued)
3
5 − 𝑥 2 − (2 − 2𝑥) 𝑑𝑥
−1
3
3 + 2𝑥 − 𝑥 2 𝑑𝑥
−1
3 3
𝑥
3𝑥 + 𝑥 2 −
3 −1
1
9 + 9 − 9 − −3 + 1 +
3
Barnett/Ziegler/Byleen Business Calculus 12e
32
=
3
11
Example 5
 Find the area bounded by
𝑓 𝑥 = 2𝑥 2 and 𝑔 𝑥 = 4 − 2𝑥
[-2, 2]
𝑓(𝑥)
𝑔(𝑥)
1
2
𝑔 𝑥 − 𝑓(𝑥) 𝑑𝑥 +
−2
𝑓 𝑥 − 𝑔(𝑥) 𝑑𝑥
1
1
2
4 − 2𝑥 − 2𝑥 2 𝑑𝑥 +
2𝑥 2 − (4 − 2𝑥) 𝑑𝑥
−2
3 1
2𝑥
4𝑥 − 𝑥 2 −
+
3 −2
1
2𝑥 3
− 4𝑥 + 𝑥 2
3
Barnett/Ziegler/Byleen Business Calculus 12e
2
1
38
=
3
12
Example 6
 Find the area bounded by
𝑦 = 25 − 𝑥 2 and y = 0; -5  x  5
5
25 − 𝑥 2 𝑑𝑥
𝐴𝑟𝑒𝑎 =
−5
1 2
𝐴𝑟𝑒𝑎 = 𝜋𝑟
2
1 2
= 𝜋5
2
25
=
𝜋
2
You don’t know how to evaluate
these types of integrals yet,
so use geometry.
Barnett/Ziegler/Byleen Business Calculus 12e
13
Homework
Barnett/Ziegler/Byleen Business Calculus 12e
14