business calculus math 1431 - lsu mathematicsdavidson/m1550/unit6_3.pdf · section 6.3 area and the...

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

IntroductionIntroduction


Introduction



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.



Introduction


a b

f



Introduction


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.

The main ideaRectanglesThe approximationLeft, Right, andMidpointThe LimitExampleExampleA comparison


The main idea



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



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.



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).



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.





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


R(f, 2, SM ) = 1.5938R(f, 2, SM ) = 5.4375





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


R(f, 3, SM ) = 1.2500R(f, 3, SM ) = 2.5000R(f, 3, SM ) = 5.7500





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


R(f, 4, SM ) = 1.0430R(f, 4, SM ) = 1.8047R(f, 4, SM ) = 3.1289R(f, 4, SM ) = 5.8594





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


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





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


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





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


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





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


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





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


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





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


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





limn→∞

R(f, n, S).


f(x) = (x−2)2+1 x ∈ [1, 4].


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



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).



Example


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.



Example


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



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 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



Example

iceExample 2: For the function f(x) = x2 + 1 on theinterval [0, 2] find R(f, 4, SR).



Example


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.



Example


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



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 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



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:



A comparison


R(f, 2, SL) = 1.0000R(f, 2, SL) = 3.0000 R(f, 2, SR) = 2.0000R(f, 2, SR) = 7.0000



A comparison


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



A comparison


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



A comparison


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



A comparison


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



A comparison


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



A comparison


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



A comparison


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



A comparison


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



A comparison


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



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.

Definite IntegralDefinite IntegralThe Main theoremGeometricInterpretation


Definite Integral



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.



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.



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 .



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



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



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




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

SummarySummaryICE


Summary

SummarySummaryICE


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.

SummarySummaryICE


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

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