Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the...
Transcript of Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the...
![Page 1: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/1.jpg)
Section 6.3 Area and the Definite Integral Business Calculus - p. 1/24
Business CalculusMath 1431
Unit 6.3Area and the Definite Integral
Mathematics Department
Louisiana State University
![Page 2: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/2.jpg)
IntroductionIntroduction
Section 6.3 Area and the Definite Integral Business Calculus - p. 2/24
Introduction
![Page 3: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/3.jpg)
IntroductionIntroduction
Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24
Introduction
As mentioned in the introductionto this chapter the core of inte-gral calculus is centered on thearea problem: Given a nonnega-tive function y = f(x) find the areaof the region bounded by the curvey = f(x), x = a, x = b, and the x-axis.
![Page 4: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/4.jpg)
IntroductionIntroduction
Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24
Introduction
As mentioned in the introductionto this chapter the core of inte-gral calculus is centered on thearea problem: Given a nonnega-tive function y = f(x) find the areaof the region bounded by the curvey = f(x), x = a, x = b, and the x-axis.
a b
f
![Page 5: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/5.jpg)
IntroductionIntroduction
Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24
Introduction
As mentioned in the introductionto this chapter the core of inte-gral calculus is centered on thearea problem: Given a nonnega-tive function y = f(x) find the areaof the region bounded by the curvey = f(x), x = a, x = b, and the x-axis.
a b
f
If the shape of this region were a rectangle wewould have no problem computing its area: lengthtimes width. However, the variety of shapes thatcan arise is as endless as there are functions f .Yet, it is through the familiar rectangle that we willapproximate the area and after a limiting processobtain the area.
![Page 6: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/6.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 4/24
The main idea
![Page 7: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/7.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 5/24
Rectangles
The main idea is to divide up the region into a se-ries of n rectangles. For convenience, each rec-tangle will have equal widths, which we will denoteby ∆x and the base of each rectangle will be aninterval on the x-axis.The height of each rectanglewill be determined by the valueof the function, f(x∗
i), for a
point x∗
iselected in the base
of the ith rectangle. Considerthe picture where we have di-vided the region into 6 rectan-gles. The height is determinedby the value of the function ateach midpoint of the base of therectangle (the dotted line). x∗
1x∗
2x∗
3x∗
4x∗
5x∗
6
![Page 8: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/8.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 6/24
The approximation
The width of each rectangle is given by ∆x = b−a
n,
the length of the interval [a, b] divided by the num-ber of rectangles. The height of the ith rectangle isf(x∗
i). Therefore its area is
f(x∗
i)∆x.
Now the sum of the areas of these rectangles is
f(x∗
1)∆x + · · · + f(x∗
n)∆x
and approximates the area of the region under f .This kind of sum is called a Riemann Sum andis defined for any function f defined on the interval[a, b]. We shall denote this sum by R(f, n, S) whereS = {x∗
1, . . . , x∗
n} is the set of selected points. Thus
R(f, n, S) = f(x∗
1)∆x + · · · + f(x∗
n)∆x.
![Page 9: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/9.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 7/24
Left, Right, and Midpoint
Remember, there are lots of ways to select thepoints S for each subdivision. If we choose themidpoints, we will denote the Riemann Sum by
R(f, n, SM ).
If we choose the left hand endpoints we will denotethe Riemann sum by
R(f, n, SL).
If we choose the right hand endpoints we will de-note the Riemann sum by
R(f, n, SR).
![Page 10: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/10.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
![Page 11: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/11.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 2, SM ) = 1.5938R(f, 2, SM ) = 5.4375
![Page 12: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/12.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 3, SM ) = 1.2500R(f, 3, SM ) = 2.5000R(f, 3, SM ) = 5.7500
![Page 13: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/13.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 4, SM ) = 1.0430R(f, 4, SM ) = 1.8047R(f, 4, SM ) = 3.1289R(f, 4, SM ) = 5.8594
![Page 14: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/14.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 5, SM ) = 0.8940R(f, 5, SM ) = 1.5000R(f, 5, SM ) = 2.2500R(f, 5, SM ) = 3.5760R(f, 5, SM ) = 5.9100
![Page 15: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/15.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 6, SM ) = 0.7813R(f, 6, SM ) = 1.3125R(f, 6, SM ) = 1.8438R(f, 6, SM ) = 2.6250R(f, 6, SM ) = 3.9063R(f, 6, SM ) = 5.9375
![Page 16: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/16.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 7, SM ) = 0.6931R(f, 7, SM ) = 1.1764R(f, 7, SM ) = 1.6071R(f, 7, SM ) = 2.1429R(f, 7, SM ) = 2.9410R(f, 7, SM ) = 4.1589R(f, 7, SM ) = 5.9541
![Page 17: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/17.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 8, SM ) = 0.6226R(f, 8, SM ) = 1.0693R(f, 8, SM ) = 1.4458R(f, 8, SM ) = 1.8574R(f, 8, SM ) = 2.4097R(f, 8, SM ) = 3.2080R(f, 8, SM ) = 4.3579R(f, 8, SM ) = 5.9648
![Page 18: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/18.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 9, SM ) = 0.5648R(f, 9, SM ) = 0.9815R(f, 9, SM ) = 1.3241R(f, 9, SM ) = 1.6667R(f, 9, SM ) = 2.0833R(f, 9, SM ) = 2.6481R(f, 9, SM ) = 3.4352R(f, 9, SM ) = 4.5185R(f, 9, SM ) = 5.9722
![Page 19: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/19.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 10, SM ) = 0.5168R(f, 10, SM ) = 0.9075R(f, 10, SM ) = 1.2263R(f, 10, SM ) = 1.5270R(f, 10, SM ) = 1.8638R(f, 10, SM ) = 2.2905R(f, 10, SM ) = 2.8613R(f, 10, SM ) = 3.6300R(f, 10, SM ) = 4.6508R(f, 10, SM ) = 5.9775
![Page 20: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/20.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 8/24
The limiting process
We now take the limit ofthese Riemann Sums:
limn→∞
R(f, n, S).
If this limit exists we callit the area of the region.To the right is an animationthat illustrates this approxi-mation and limiting process.The function is
f(x) = (x−2)2+1 x ∈ [1, 4].
In the blue box is the Rie-mann sum for various val-ues of n.
R(f, 11, SM ) = 0.4761R(f, 11, SM ) = 0.8441R(f, 11, SM ) = 1.1444R(f, 11, SM ) = 1.4177R(f, 11, SM ) = 1.7045R(f, 11, SM ) = 2.0455R(f, 11, SM ) = 2.4810R(f, 11, SM ) = 3.0518R(f, 11, SM ) = 3.7985R(f, 11, SM ) = 4.7615R(f, 11, SM ) = 5.9814
![Page 21: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/21.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 9/24
Example
Example 1: Let f(x) = x2 + 1 on the interval [0, 2].Find the Riemann Sum for f using n = 4 subdivi-sions and choosing the left hand endpoints in eachsubdivision. I.e. compute R(f, 4, SL).
![Page 22: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/22.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 9/24
Example
Example 1: Let f(x) = x2 + 1 on the interval [0, 2].Find the Riemann Sum for f using n = 4 subdivi-sions and choosing the left hand endpoints in eachsubdivision. I.e. compute R(f, 4, SL).
In this case ∆x = b−a
n= 2
4= .5, the subdivisions
are[0, .5] [.5, 1] [1, 1.5] [1.5, 2],
and the left hand endpoints are
x∗
1= 0, x∗
2= .5, x∗
3= 1, x∗
4= 1.5.
![Page 23: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/23.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 9/24
Example
Example 1: Let f(x) = x2 + 1 on the interval [0, 2].Find the Riemann Sum for f using n = 4 subdivi-sions and choosing the left hand endpoints in eachsubdivision. I.e. compute R(f, 4, SL).
In this case ∆x = b−a
n= 2
4= .5, the subdivisions
are[0, .5] [.5, 1] [1, 1.5] [1.5, 2],
and the left hand endpoints are
x∗
1= 0, x∗
2= .5, x∗
3= 1, x∗
4= 1.5.
The following table gives the value of f at each ofthese points:
x 0 .5 1 1.5
x2 + 1 1 1.25 2 3.25
![Page 24: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/24.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 10/24
The Riemann Sum
We now get
R(f, 4, SL) = f(x∗
1)∆x + f(x∗
2)∆x + f(x∗
3)∆x + f(x∗
4)∆x
= 1(.5) + 1.25(.5) + 2(.5) + 3.25(.5)
= 3.75
![Page 25: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/25.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 10/24
The Riemann Sum
We now get
R(f, 4, SL) = f(x∗
1)∆x + f(x∗
2)∆x + f(x∗
3)∆x + f(x∗
4)∆x
= 1(.5) + 1.25(.5) + 2(.5) + 3.25(.5)
= 3.75
The picture below illus-trates the left Riemannsum.
1.00
1.25
2.00
3.25
0 .5 1 1.5 2
![Page 26: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/26.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 11/24
Example
iceExample 2: For the function f(x) = x2 + 1 on theinterval [0, 2] find R(f, 4, SR).
![Page 27: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/27.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 11/24
Example
iceExample 2: For the function f(x) = x2 + 1 on theinterval [0, 2] find R(f, 4, SR).
This is the right Riemann sum for the function inexample 1. The subdivisions are the same
[0, .5] [.5, 1] [1, 1.5] [1.5, 2]
but the selected points are
x∗
1= .5 x∗
2= 1 x∗
3= 1.5 x∗
4= 2.
![Page 28: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/28.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 11/24
Example
iceExample 2: For the function f(x) = x2 + 1 on theinterval [0, 2] find R(f, 4, SR).
This is the right Riemann sum for the function inexample 1. The subdivisions are the same
[0, .5] [.5, 1] [1, 1.5] [1.5, 2]
but the selected points are
x∗
1= .5 x∗
2= 1 x∗
3= 1.5 x∗
4= 2.
Again the table gives the relevant data:
x .5 1 1.5 2
x2 + 1 1.25 2 3.25 5
![Page 29: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/29.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 12/24
The Riemann Sum
We now get
R(f, 4, SR) = f(x∗
1)∆x + f(x∗
2)∆x + f(x∗
3)∆x + f(x∗
4)∆x
= 1.25(.5) + 2(.5) + 3.25(.5) + 5(.5)
= 5.75
![Page 30: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/30.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 12/24
The Riemann Sum
We now get
R(f, 4, SR) = f(x∗
1)∆x + f(x∗
2)∆x + f(x∗
3)∆x + f(x∗
4)∆x
= 1.25(.5) + 2(.5) + 3.25(.5) + 5(.5)
= 5.75
The picture below il-lustrates the right Rie-mann sum.
1.25
2.00
3.25
5.00
0 .5 1 1.5 2
![Page 31: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/31.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
![Page 32: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/32.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 2, SL) = 1.0000R(f, 2, SL) = 3.0000 R(f, 2, SR) = 2.0000R(f, 2, SR) = 7.0000
![Page 33: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/33.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 3, SL) = 0.6667R(f, 3, SL) = 1.6296R(f, 3, SL) = 3.4815 R(f, 3, SR) = 0.9630R(f, 3, SR) = 2.8148R(f, 3, SR) = 6.1481
![Page 34: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/34.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 4, SL) = 0.5000R(f, 4, SL) = 1.1250R(f, 4, SL) = 2.1250R(f, 4, SL) = 3.7500 R(f, 4, SR) = 0.6250R(f, 4, SR) = 1.6250R(f, 4, SR) = 3.2500R(f, 4, SR) = 5.7500
![Page 35: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/35.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 5, SL) = 0.4000R(f, 5, SL) = 0.8640R(f, 5, SL) = 1.5200R(f, 5, SL) = 2.4960R(f, 5, SL) = 3.9200 R(f, 5, SR) = 0.4640R(f, 5, SR) = 1.1200R(f, 5, SR) = 2.0960R(f, 5, SR) = 3.5200R(f, 5, SR) = 5.5200
![Page 36: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/36.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 6, SL) = 0.3333R(f, 6, SL) = 0.7037R(f, 6, SL) = 1.1852R(f, 6, SL) = 1.8519R(f, 6, SL) = 2.7778R(f, 6, SL) = 4.0370 R(f, 6, SR) = 0.3704R(f, 6, SR) = 0.8519R(f, 6, SR) = 1.5185R(f, 6, SR) = 2.4444R(f, 6, SR) = 3.7037R(f, 6, SR) = 5.3704
![Page 37: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/37.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 7, SL) = 0.2857R(f, 7, SL) = 0.5948R(f, 7, SL) = 0.9738R(f, 7, SL) = 1.4694R(f, 7, SL) = 2.1283R(f, 7, SL) = 2.9971R(f, 7, SL) = 4.1224 R(f, 7, SR) = 0.3090R(f, 7, SR) = 0.6880R(f, 7, SR) = 1.1837R(f, 7, SR) = 1.8426R(f, 7, SR) = 2.7114R(f, 7, SR) = 3.8367R(f, 7, SR) = 5.2653
![Page 38: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/38.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 8, SL) = 0.2500R(f, 8, SL) = 0.5156R(f, 8, SL) = 0.8281R(f, 8, SL) = 1.2188R(f, 8, SL) = 1.7188R(f, 8, SL) = 2.3594R(f, 8, SL) = 3.1719R(f, 8, SL) = 4.1875 R(f, 8, SR) = 0.2656R(f, 8, SR) = 0.5781R(f, 8, SR) = 0.9688R(f, 8, SR) = 1.4688R(f, 8, SR) = 2.1094R(f, 8, SR) = 2.9219R(f, 8, SR) = 3.9375R(f, 8, SR) = 5.1875
![Page 39: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/39.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 9, SL) = 0.2222R(f, 9, SL) = 0.4554R(f, 9, SL) = 0.7215R(f, 9, SL) = 1.0425R(f, 9, SL) = 1.4403R(f, 9, SL) = 1.9369R(f, 9, SL) = 2.5542R(f, 9, SL) = 3.3141R(f, 9, SL) = 4.2387 R(f, 9, SR) = 0.2332R(f, 9, SR) = 0.4993R(f, 9, SR) = 0.8203R(f, 9, SR) = 1.2181R(f, 9, SR) = 1.7147R(f, 9, SR) = 2.3320R(f, 9, SR) = 3.0919R(f, 9, SR) = 4.0165R(f, 9, SR) = 5.1276
![Page 40: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/40.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 10, SL) = 0.2000R(f, 10, SL) = 0.4080R(f, 10, SL) = 0.6400R(f, 10, SL) = 0.9120R(f, 10, SL) = 1.2400R(f, 10, SL) = 1.6400R(f, 10, SL) = 2.1280R(f, 10, SL) = 2.7200R(f, 10, SL) = 3.4320R(f, 10, SL) = 4.2800 R(f, 10, SR) = 0.2080R(f, 10, SR) = 0.4400R(f, 10, SR) = 0.7120R(f, 10, SR) = 1.0400R(f, 10, SR) = 1.4400R(f, 10, SR) = 1.9280R(f, 10, SR) = 2.5200R(f, 10, SR) = 3.2320R(f, 10, SR) = 4.0800R(f, 10, SR) = 5.0800
![Page 41: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/41.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 13/24
A comparison
The last two examples illustrate that the Riemannsum is dependent on the way we select the pointsin the subdivision. However, for large numbers ofsubdivisions the left and right Riemann Sum willhave values close. We observe this in the two pic-tures below:
R(f, 11, SL) = 0.1818R(f, 11, SL) = 0.3696R(f, 11, SL) = 0.5755R(f, 11, SL) = 0.8114R(f, 11, SL) = 1.0894R(f, 11, SL) = 1.4215R(f, 11, SL) = 1.8197R(f, 11, SL) = 2.2960R(f, 11, SL) = 2.8625R(f, 11, SL) = 3.5312R(f, 11, SL) = 4.3140 R(f, 11, SR) = 0.1878R(f, 11, SR) = 0.3937R(f, 11, SR) = 0.6296R(f, 11, SR) = 0.9076R(f, 11, SR) = 1.2397R(f, 11, SR) = 1.6379R(f, 11, SR) = 2.1142R(f, 11, SR) = 2.6807R(f, 11, SR) = 3.3494R(f, 11, SR) = 4.1322R(f, 11, SR) = 5.0413
![Page 42: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/42.jpg)
The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison
Section 6.3 Area and the Definite Integral Business Calculus - p. 14/24
f(x) = x2 + 1 [0, 2]
In the table below we have calculated the left andright Riemann sum for large values of n.
n 5 10 100 1000 10000 100000
R(f, n, SL) 3.92 4.28 4.6268 4.6627 4.6662 4.6666
R(f, n, SR) 5.52 5.08 4.7068 4.6707 4.6670 4.6667
A reasonable conjecture is that the left and rightRiemann sums for x2 + 1 on [a, b] converges to4.6666̄ = 42
3. We will show that this is indeed the
case in the next section, section 6.4.
![Page 43: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/43.jpg)
Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation
Section 6.3 Area and the Definite Integral Business Calculus - p. 15/24
Definite Integral
![Page 44: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/44.jpg)
Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation
Section 6.3 Area and the Definite Integral Business Calculus - p. 16/24
The Definite Integral
For f any function (whether positive or negative)on an interval [a, b] the Riemann sum
R(f, n, S) = f(x∗
1)∆x + · · · + f(x∗
n)∆x
is well defined. If the limit
limn→∞
R(f, n, S)
exists, regardless of how the x∗
i’s are chosen, then
we will write the limit∫
b
a
f(x) dx
and we call this limit the definite integral of f
between a and b. We will say f is integrable . Wecall the number a the lower limit of integrationand b the upper limit of integration.
![Page 45: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/45.jpg)
Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation
Section 6.3 Area and the Definite Integral Business Calculus - p. 17/24
The Main theorem
Theorem: Let f be a continuous function on aclosed interval [a, b]. Then f is integrable. Ie.
∫b
a
f(x) dx
exists as a limit of Riemann Sums.
![Page 46: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/46.jpg)
Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation
Section 6.3 Area and the Definite Integral Business Calculus - p. 17/24
The Main theorem
Theorem: Let f be a continuous function on aclosed interval [a, b]. Then f is integrable. Ie.
∫b
a
f(x) dx
exists as a limit of Riemann Sums.
Determining the limit of a Riemann sum can be anightmare of calculations. In the next section, wewill discuss the fundamental theorems of calculusthat make calculating Riemann sums unnecessarywhen we can find an antiderivative of f .
![Page 47: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/47.jpg)
Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation
Section 6.3 Area and the Definite Integral Business Calculus - p. 18/24
The Geometric Interpretation: f ≥ 0
Suppose f is a nonnegative continuous functionon the interval [a, b]. As our discussion has re-vealed we interpret
∫b
af(x) dx as the area of the re-
gion bounded by the graph of y = f(x), a ≤ x ≤ b,the x-axis, and the vertical lines x = a and x = b asillustrated below:
Area =Rb
af(x) dx
![Page 48: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/48.jpg)
Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation
Section 6.3 Area and the Definite Integral Business Calculus - p. 19/24
General f
Now suppose f is a continuous function on aclosed interval [a,b]. Here we allow f to take onnegative values. The definition of the Riemannsum still makes sense but its interpretation as anapproximation to area needs to be somewhat ad-justed. Consider the graph of the function below.There are two regions that are bounded by the x-axis: one below and one above.
a b
f
![Page 49: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/49.jpg)
Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation
Section 6.3 Area and the Definite Integral Business Calculus - p. 20/24
A typical Riemann sum
Riemann Sums
R(f, n, S) = f(x∗
1)∆x + · · · + f(x∗
n)∆x
make perfectly good sense for such functions.However, when x∗
iis in an interval where f is neg-
ative the corresponding product f(x∗
i)∆x is nega-
tive. This is illustrated in the picture below wheref(x∗
i)∆x is negative the area of the rectangle be-
low the x-axis.
a b
f
![Page 50: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/50.jpg)
Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation
Section 6.3 Area and the Definite Integral Business Calculus - p. 21/24
The limiting process
After taking limits we are lead to the following re-sult:
If f is continuous on [a, b] then∫
b
a
f(x) dx
is equal to the area of the region R1 above the x-axis minus the area of the region R2 below.
R2
R1
a b
f
![Page 51: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/51.jpg)
SummarySummaryICE
Section 6.3 Area and the Definite Integral Business Calculus - p. 22/24
Summary
![Page 52: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/52.jpg)
SummarySummaryICE
Section 6.3 Area and the Definite Integral Business Calculus - p. 23/24
Summary
Here are some of the key concepts you shouldunderstand:
■ The Riemann sum R(f, n, S) and how it is usedto approximate area.
■ The special Riemann sums R(f, n, SL),R(f, n, SR), and R(f, n, SM ).
■ The main theorem: A continuous function on aclosed interval has Riemann sums thatconverge.
■ The notation∫
b
af(x) dx.
■ The geometric interpretation of∫
b
af(x) dx.
![Page 53: Business Calculus Math 1431 - LSU Mathematicsdavidson/m1550/Unit6_3.pdf · Section 6.3 Area and the Definite Integral Business Calculus - p. 3/24 Introduction As mentioned in the](https://reader030.fdocuments.net/reader030/viewer/2022040405/5e99d8ae40814707da1e20a9/html5/thumbnails/53.jpg)
SummarySummaryICE
Section 6.3 Area and the Definite Integral Business Calculus - p. 24/24
In-Class Exercise
returnIn-Class Exercise 1: Find the midpoint Riemannsum for
y = 2x + 1
on the interval [0, 4] with n = 4.
1. 20
2. 16
3. 24
4. 22
5. None of the above