C.1.3b - Area & Definite Integrals: An Algebraic Perspective Calculus - Santowski...

C.1.3b - Area & Definite Integrals: An Algebraic Perspective

Calculus - Santowski

04/21/23Calculus - Santowski1

Fast Five

04/21/23Calculus - Santowski2

(1) 0.25[f(1) + f(1.25) + f(1.5) + f(1.75)] if f(x)

= x2 + x - 1

(2) Illustrate Q1 with a diagram, showing all

relevant details

(A) Review

04/21/23Calculus - Santowski3

We will continue to move onto a second type of integral the definite integral

Last lesson, we estimated the area under a curve by constructing/drawing rectangles under the curve

Today, we will focus on doing the same process, but from an algebraic perspective, using summations

(B) Summations, ∑

04/21/23Calculus - Santowski4

A series is the sum of a sequence (where a sequence is simply a list of numbers)

Ex: the sequence 2,4,6,8,10,12, …..2n has a an associated sum, written as:

The sum 2 + 4 + 6 + 8 + … + 2n can also be written as 2(1+ 2 + 3 + 4 + … + n) or:€

Sn = 2 + 4 + 6 + 8 + ...+ 2n = 2ii=1

n

∑

€

Sn = 2 1+ 2 + 3+ 4 + ...+ n( ) = 2 × ii=1

n

∑

(C) Working with Summations

04/21/23Calculus - Santowski5

Ex 1. You are given the following series:

List the first 7 terms of each series

€

(i) i + 3i=1

20

∑

(ii) i3

i= 41

100

∑

(iii) 3i2 −1( )i=1

n

∑

(iv) 2i3 + 3i − 2( )i=1

n

∑

(C) Working with Summations

04/21/23Calculus - Santowski6

Ex 1I. You are given the following series:

Evaluate each series

€

(i) i + 3i=1

20

∑

(ii) i3

i=1

5

∑

(iii) 3i2 −1( )i=1

6

∑

(iv) 2i3 + 3i − 2( )i= 0

4

∑

(C) Working with Summations

04/21/23Calculus - Santowski7

Ex 1. You are given the following series:

List the first 10 terms of each series

€

(i) ii=1

n

∑

(ii) i2

i= 41

n

∑

(iii) i3

i=1

n

∑

(C) Working with Summations

04/21/23Calculus - Santowski8

Ex 1. You are given the following series:

Evaluate each series

€

(i) ii=1

10

∑

(ii) i2

i= 41

7

∑

(iii) i3

i=1

4

∑

(B) Summations, ∑

04/21/23Calculus - Santowski9

Here are “sum” important summation formulas:

(I) sum of the natural numbers (1+2+3+4+….n)

(II) sum of squares (1+4+9+16+….+n2)

(III) sum of cubes (1+8+27+64+….n3)

€

ii=1

n

∑ =n n +1( )

2

€

i2

i=1

n

∑ =n n +1( ) 2n +1( )

6

€

i3

i=1

n

∑ =n2 n +1( )

2

4

(E) Working With Summations - GDC

Now, to save all the tedious algebra (YEAHH!!!), let’s use the TI-89 to do sums

First, let’s confirm our summation formulas for i, i2 & i3 and get acquainted with the required syntax

04/21/23 Calculus - Santowski 10

(E) Working With Summations - GDC

So let’s revisit our previous example of

And our 4th example of

04/21/23 Calculus - Santowski 11

€

(i) i + 3i=1

20

∑

€

(iv) 2i3 + 3i − 2( )i=1

n

∑

(F) Applying Summations

04/21/23Calculus - Santowski12

So what do we need summations for?

Let’s connect this algebra skill to determining the area under curves ==> after all, we are simply summing areas of individual rectangles to estimate an area under a curve

PART 2 - The Area Problem

04/21/23Calculus - Santowski13

Let’s work with a simple quadratic function, f(x) = x2 + 2 and use a specific interval of [0,3]

Now we wish to estimate the area under this curve

(A) The Area Problem – Rectangular Approximation Method (RAM)

04/21/23Calculus - Santowski14

To estimate the area under the curve, we will divide the are into simple rectangles as we can easily find the area of rectangles A = l × w

Each rectangle will have a width of x which we calculate as (b – a)/n where b represents the higher bound on the area (i.e. x = 3) and a represents the lower bound on the area (i.e. x = 0) and n represents the number of rectangles we want to construct

The height of each rectangle is then simply calculated using the function equation

Then the total area (as an estimate) is determined as we sum the areas of the numerous rectangles we have created under the curve

AT = A1 + A2 + A3 + ….. + An

We can visualize the process on the next slide

(A) The Area Problem – Rectangular Approximation Method (RRAM)

04/21/23Calculus - Santowski15

We have chosen to draw 6 rectangles on the interval [0,3]

A1 = ½ × f(½) = 1.125

A2 = ½ × f(1) = 1.5

A3 = ½ × f(1½) = 2.125

A4 = ½ × f(2) = 3

A5 = ½ × f(2½) = 4.125

A6 = ½ × f(3) = 5.5

AT = 17.375 square units So our estimate is 17.375

which is obviously an overestimate

(B) Sums & Rectangular Approximation Method (RRAM)

So let’s apply our summation formulas:

Each rectangle’s area is f(xi)x where f(x) = x2 + 2

x = 0.5 and xi = 0 + xi

Therefore the area of 6 rectangles is given by

€

A = f (x i)Δxi=1

6

∑

A = f 0 + Δxi( ) ×1

2i=1

6

∑

A =1

2× f 0 +

1

2i

⎛

⎝ ⎜

⎞

⎠ ⎟

i=1

6

∑

04/21/23 Calculus - Santowski 16

(C) The Area Problem – Rectangular Approximation Method (LRAM)

04/21/23Calculus - Santowski17

In our previous slide, we used 6 rectangles which were constructed using a “right end point” (realize that both the use of 6 rectangles and the right end point are arbitrary!) in an increasing function like f(x) = x2 + 2 this creates an over-estimate of the area under the curve

So let’s change from the right end point to the left end point and see what happens

(C) The Area Problem – Rectangular Approximation Method (LRAM)

04/21/23Calculus - Santowski18

We have chosen to draw 6 rectangles on the interval [0,3]

A1 = ½ × f(0) = 1

A2 = ½ × f(½) = 1.125

A3 = ½ × f(1) = 1.5

A4 = ½ × f(1½) = 2.125

A5 = ½ × f(2) = 3

A6 = ½ × f(2½) = 4.125

AT = 12.875 square units So our estimate is 12.875

which is obviously an under-estimate

(D) Sums & Rectangular Approximation Method (LRAM) So let’s apply our

summation formulas:

Each rectangle’s area is f(xi)x where f(x) = x2 + 2

x = 0.5 and xi = 0 + xi

Therefore the area of 6 rectangles is given by: (Notice change in i???)

€

A = f (x i)Δxi= 0

5

∑

A = f 0 + Δxi( ) ×1

2i= 0

5

∑

A =1

2× f 0 +

1

2i

⎛

⎝ ⎜

⎞

⎠ ⎟

i= 0

5

∑

04/21/23 Calculus - Santowski 19

(E) The Area Problem – Rectangular Approximation Method (MRAM)

04/21/23Calculus - Santowski20

So our “left end point” method (called a left hand Riemann sum or LRAM) gives us an underestimate (in this example)

Our “right end point” method (called a right handed Riemann sum or RRAM) gives us an overestimate (in this example)

We can adjust our strategy in a variety of ways one is by adjusting the “end point” why not simply use a “midpoint” in each interval and get a mix of over- and under-estimates? see next slide

(E) The Area Problem – Rectangular Approximation Method (MRAM)

04/21/23Calculus - Santowski21

We have chosen to draw 6 rectangles on the interval [0,3]

A1 = ½ × f(¼) = 1.03125 A2 = ½ × f (¾) = 1.28125 A3 = ½ × f(1¼) = 1.78125 A4 = ½ × f(1¾) = 2.53125 A5 = ½ × f(2¼) = 3.53125 A6 = ½ × f(2¾) = 4.78125 AT = 14.9375 square units

which is a more accurate estimate (15 is the exact answer)

(F) Sums & Rectangular Approximation Method (MRAM)

So let’s apply our summation formulas:

Each rectangle’s area is f(xi)x where f(x) = x2 + 2

x = 0.5 and xi = 0.25 + xi

Therefore the area of 6 rectangles is given by: (Notice change in i??? and the xi expression???)

€

A = f (x i)Δxi= 0

5

∑

A = f1

4+ Δxi

⎛

⎝ ⎜

⎞

⎠ ⎟×

1

2i= 0

5

∑

A =1

2× f

1

4+

1

2i

⎛

⎝ ⎜

⎞

⎠ ⎟

i= 0

5

∑

04/21/23 Calculus - Santowski 22

(G) The Area Problem – Expanding our Example

04/21/23Calculus - Santowski23

Now back to our left and right Riemann sums and our original example how can we increase the accuracy of our estimate?

We simply increase the number of rectangles that we construct under the curve

Initially we chose 6, now let’s choose a few more … say 12, 60, and 300 ….

But first, we need to generalize our specific formula!

(G) The Area Problem – Expanding our Example

04/21/23Calculus - Santowski24

So we have

Now, x = (b-a)/n = (3 - 0)/n = 3/n And f(xi) = f(a + xi) = f(0 + 3i/n) = f(3i/n)

So, we have to work with the generalized formula

€

A = f x i( )i=1

n

∑ Δx

€

A = f3i

n

⎛

⎝ ⎜

⎞

⎠ ⎟

i=1

n

∑ ×3

n

(G) The Area Problem – Expanding our Example

Does this generalized formula work?

Well, test it with n = 6 as before!

€

A = f3i

n

⎛

⎝ ⎜

⎞

⎠ ⎟

i=1

n

∑ ×3

n

04/21/23 Calculus - Santowski 25

(G) The Area Problem – Expanding our Example (RRAM)

# of rectangles Area estimate

12 16.15625

60 15.22625

300 15.04505

04/21/23 Calculus - Santowski 26

(G) The Area Problem – Expanding our Example

# of rectangles Area estimate

12 16.15625

60 15.22625

300 15.04505

04/21/23 Calculus - Santowski 27

(G) The Area Problem – Expanding our Example (LRAM)

# of rectangles Area estimate

12 13.90625

60 14.77625

300 14.95505

04/21/23 Calculus - Santowski 28

(G) The Area Problem – Expanding our Example (LRAM)

# of rectangles Area estimate

12 13.90625

60 14.77625

300 14.95505

04/21/23 Calculus - Santowski 29

(H) The Area Problem - Conclusion So our exact area seems to be “sandwiched” between

14.95505 and 15.04505 !!! So, if increasing the number of rectangles increases the

accuracy, the question that needs to be asked is ….. how many rectangles should be used???

The answer is ….. why not use an infinite number of rectangles!! so now we are back into LIMITS!!

So, the exact area between the curve and the x-axis can be determined by evaluating the following limit:

04/21/23 Calculus - Santowski 30

€

A = limn→∞

f x i( )Δxi=1

n

∑

(H) The Area Problem - Conclusion

So let’s verify the example using the GDC and limits:

04/21/23 Calculus - Santowski 31

(I) The Area Problem – Further Examples

04/21/23Calculus - Santowski32

(i) Determine the area between the curve f(x) = x3 – 5x2 + 6x + 5 and the x-axis on

[0,4] if we (a) construct 20 rectangles or (b) if we want the exact area

(ii) Determine the exact area between the curve f(x) = x2 – 4 and the x-axis on [0,2] if we (a)

construct 30 rectangles or (b) if we want the exact area

(iii) Determine the exact area between the curve f(x) = x2 – 2 and the x-axis on [0,2] if we (a)

construct 10 rectangles or (b) if we want the exact area

(J) Homework

04/21/23Calculus - Santowski33

Homework to reinforce these concepts from this second part of our lesson:

Handout, Stewart, Calculus - A First Course, 1989, Exercise 10.4, p474-5, Q3,4,6a

Internet Links

04/21/23Calculus - Santowski34

Calculus I (Math 2413) - Integrals - Area Problem from Paul Dawkins

Integration Concepts from Visual Calculus

Areas and Riemann Sums from P.K. Ving - Calculus I - Problems and Solutions