C.1.3b - Area & Definite Integrals: An Algebraic Perspective
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C.1.3b - Area & Definite Integrals: An Algebraic Perspective
Calculus - Santowski
04/24/23Calculus - Santowski 1
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Fast Five
04/24/23Calculus - 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
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(A) Review
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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
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(B) Summations, ∑
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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
∑
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(C) Working with Summations
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Ex 1. You are given the following series:
List the first 7 terms of each series
€
(i) i + 3i=1
20
∑
(ii) i3i= 41
100
∑
(iii) 3i2 −1( )i=1
n
∑
(iv) 2i3 + 3i − 2( )i=1
n
∑
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(C) Working with Summations
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Ex 1I. You are given the following series:
Evaluate each series
€
(i) i + 3i=1
20
∑
(ii) i3i=1
5
∑
(iii) 3i2 −1( )i=1
6
∑
(iv) 2i3 + 3i − 2( )i= 0
4
∑
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(C) Working with Summations
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Ex 1. You are given the following series:
List the first 10 terms of each series
€
(i) ii=1
n
∑
(ii) i2i= 41
n
∑
(iii) i3i=1
n
∑
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(C) Working with Summations
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Ex 1. You are given the following series:
Evaluate each series
€
(i) ii=1
10
∑
(ii) i2i= 41
7
∑
(iii) i3i=1
4
∑
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(B) Summations, ∑
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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
€
i2i=1
n
∑ =n n +1( ) 2n +1( )
6
€
i3i=1
n
∑ =n2 n +1( )
2
4
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(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
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(E) Working With Summations - GDC So let’s revisit our
previous example of
And our 4th example of
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€
(i) i + 3i=1
20
∑
€
(iv) 2i3 + 3i − 2( )i=1
n
∑
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(F) Applying Summations
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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
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PART 2 - The Area Problem
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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
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(A) The Area Problem – Rectangular Approximation Method (RAM)
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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
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(A) The Area Problem – Rectangular Approximation Method (RRAM)
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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
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(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( ) × 12i=1
6
∑
A = 12
× f 0 + 12i
⎛ ⎝ ⎜
⎞ ⎠ ⎟
i=1
6
∑
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(C) The Area Problem – Rectangular Approximation Method (LRAM)
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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
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(C) The Area Problem – Rectangular Approximation Method (LRAM)
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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
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(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( ) × 12i= 0
5
∑
A = 12
× f 0 + 12i
⎛ ⎝ ⎜
⎞ ⎠ ⎟
i= 0
5
∑
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(E) The Area Problem – Rectangular Approximation Method (MRAM)
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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
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(E) The Area Problem – Rectangular Approximation Method (MRAM)
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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)
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(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 = f 14
+ Δxi ⎛ ⎝ ⎜
⎞ ⎠ ⎟× 1
2i= 0
5
∑
A = 12
× f 14
+ 12i
⎛ ⎝ ⎜
⎞ ⎠ ⎟
i= 0
5
∑
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(G) The Area Problem – Expanding our Example
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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!
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(G) The Area Problem – Expanding our Example
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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 = f 3in ⎛ ⎝ ⎜
⎞ ⎠ ⎟
i=1
n
∑ × 3n
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(G) The Area Problem – Expanding our Example
Does this generalized formula work?
Well, test it with n = 6 as before!
€
A = f 3in ⎛ ⎝ ⎜
⎞ ⎠ ⎟
i=1
n
∑ × 3n
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(G) The Area Problem – Expanding our Example (RRAM)
# of rectangles Area estimate
12 16.15625
60 15.22625
300 15.04505
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(G) The Area Problem – Expanding our Example
# of rectangles Area estimate
12 16.15625
60 15.22625
300 15.04505
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(G) The Area Problem – Expanding our Example (LRAM)
# of rectangles Area estimate
12 13.90625
60 14.77625
300 14.95505
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(G) The Area Problem – Expanding our Example (LRAM)
# of rectangles Area estimate
12 13.90625
60 14.77625
300 14.95505
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(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:
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€
A = limn→∞
f x i( )Δxi=1
n
∑
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(H) The Area Problem - Conclusion
So let’s verify the example using the GDC and limits:
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(I) The Area Problem – Further Examples
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(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
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(J) Homework
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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
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Internet Links
04/24/23Calculus - 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