The document discusses calculating the area of a region R. It introduces using a ruler x to measure the span of R from x=a to x=b. It defines the cross-sectional length L(x) and partitions the interval [a,b] into subintervals. The Riemann sum of the areas of approximating rectangles is shown to approach the actual area of R, defined as the definite integral of L(x) from a to b. As an example, it calculates the area between the curves y=-x^2+2x and y=x^2 by finding the interval spans from 0 to 1 and taking the integral of the difference of the functions.
State of ICS and IoT Cyber Threat Landscape Report 2024 previewPrayukth K V
The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
Transcript: Selling digital books in 2024: Insights from industry leaders - T...BookNet Canada
The publishing industry has been selling digital audiobooks and ebooks for over a decade and has found its groove. What’s changed? What has stayed the same? Where do we go from here? Join a group of leading sales peers from across the industry for a conversation about the lessons learned since the popularization of digital books, best practices, digital book supply chain management, and more.
Link to video recording: https://bnctechforum.ca/sessions/selling-digital-books-in-2024-insights-from-industry-leaders/
Presented by BookNet Canada on May 28, 2024, with support from the Department of Canadian Heritage.
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
Key Trends Shaping the Future of Infrastructure.pdfCheryl Hung
Keynote at DIGIT West Expo, Glasgow on 29 May 2024.
Cheryl Hung, ochery.com
Sr Director, Infrastructure Ecosystem, Arm.
The key trends across hardware, cloud and open-source; exploring how these areas are likely to mature and develop over the short and long-term, and then considering how organisations can position themselves to adapt and thrive.
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
Let's dive deeper into the world of ODC! Ricardo Alves (OutSystems) will join us to tell all about the new Data Fabric. After that, Sezen de Bruijn (OutSystems) will get into the details on how to best design a sturdy architecture within ODC.
LF Energy Webinar: Electrical Grid Modelling and Simulation Through PowSyBl -...DanBrown980551
Do you want to learn how to model and simulate an electrical network from scratch in under an hour?
Then welcome to this PowSyBl workshop, hosted by Rte, the French Transmission System Operator (TSO)!
During the webinar, you will discover the PowSyBl ecosystem as well as handle and study an electrical network through an interactive Python notebook.
PowSyBl is an open source project hosted by LF Energy, which offers a comprehensive set of features for electrical grid modelling and simulation. Among other advanced features, PowSyBl provides:
- A fully editable and extendable library for grid component modelling;
- Visualization tools to display your network;
- Grid simulation tools, such as power flows, security analyses (with or without remedial actions) and sensitivity analyses;
The framework is mostly written in Java, with a Python binding so that Python developers can access PowSyBl functionalities as well.
What you will learn during the webinar:
- For beginners: discover PowSyBl's functionalities through a quick general presentation and the notebook, without needing any expert coding skills;
- For advanced developers: master the skills to efficiently apply PowSyBl functionalities to your real-world scenarios.
The Art of the Pitch: WordPress Relationships and SalesLaura Byrne
Clients don’t know what they don’t know. What web solutions are right for them? How does WordPress come into the picture? How do you make sure you understand scope and timeline? What do you do if sometime changes?
All these questions and more will be explored as we talk about matching clients’ needs with what your agency offers without pulling teeth or pulling your hair out. Practical tips, and strategies for successful relationship building that leads to closing the deal.
Kubernetes & AI - Beauty and the Beast !?! @KCD Istanbul 2024Tobias Schneck
As AI technology is pushing into IT I was wondering myself, as an “infrastructure container kubernetes guy”, how get this fancy AI technology get managed from an infrastructure operational view? Is it possible to apply our lovely cloud native principals as well? What benefit’s both technologies could bring to each other?
Let me take this questions and provide you a short journey through existing deployment models and use cases for AI software. On practical examples, we discuss what cloud/on-premise strategy we may need for applying it to our own infrastructure to get it to work from an enterprise perspective. I want to give an overview about infrastructure requirements and technologies, what could be beneficial or limiting your AI use cases in an enterprise environment. An interactive Demo will give you some insides, what approaches I got already working for real.
2. We are to find the area of a given enclosed region R.
R
More on Areas
3. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
R
More on Areas
x
4. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
R
a=x b=x
x
5. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
R
a=x b=x
x
6. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
R
L(x)
a=x b=x
x
x
7. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b]
R
a=x b=x
x
L(x)
x
8. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b]
x1 x2 xi–1 xi
R
a=x0 b=xn
x
9. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x, x, x, .. x=b} be a regular partition of
012n[a, b] and select an arbitrary point x*
in each sub-interval
i [x , x].
i–1ix1 x2 xi–1 xi
R
a=x0 b=xn
x
10. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x, x, x, .. x=b} be a regular partition of
012n[a, b] and select an arbitrary point x*
in each sub-interval
i [x , x].
i–1ia=xxxb=x
0 1 2
x1 *
xi–1 xi
R
More on Areas
x
11. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi].
x1 * x2 *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
*
x
12. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi].
x1 * x2 * x3 *
xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
*
x
13. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi].
x1 * x2 * x3 *
xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
L(xi)
*
*
x
14. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi].
x1 * x2 * x3 *
xi *
cross–sectional
length at xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
L(xi)
*
*
x
15. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi]. Let Δx = the width of each subinterval.
x1 * x2 * x3 *
xi *
cross–sectional
length at xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
*
L(xi*)
x
16. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi]. Let Δx = the width of each subinterval. The rectangle
with L(xi) as height and Δx as width approximates the area in R
that is spanned from xi–1 to xi.
x1 * x2 * x3 *
xi *
cross–sectional
length at xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
*
*
Δx
L(xi*)
x
17. More on Areas
R
x1 * x2 * x3 * xi *
a=x0 x b=x 1 x2 xi–1 xi
The Riemann sum
x
18. More on Areas
R
Δx
L(x1*)
x1 * x2 * x3 * xi *
a=x0 x b=x 1 x2 xi–1 xi
L(x1The Riemann sum *)Δx
x
19. More on Areas
R
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
a=x0 x b=x 1 x2 xi–1 xi
L(x1*)Δx+ L(x2The Riemann sum *)Δx+ …
x
20. More on Areas
Δx
*
L(xi)
R
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
Δx
L(xn* )
xn* x
a=x0 x b=x 1 x2 xi–1 xi
L(x1*)Δx+ L(x2*)Δx+ … L(xnThe Riemann sum *)Δx
21. More on Areas
Δx
L(xi)
R
x1 * x2 * x3 * xi *
Δx
L(xn* )
a=x0 x b=x 1 x2 xi–1 xi
n
The Riemann sum L(x*)Δx+
L(x *)Δx+ … L(x*)Δx =
Σ L(x*)Δx 12nii=1
of all such rectangles approximates the area of R.
L(x1) L(x2* *)
*
Δx
Δx
xn*
x
22. More on Areas
Δx
*
L(xi)
R
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
Δx
L(xn* )
xn*
a=x0 x b=x 1 x2 xi–1 xi
n
The Riemann sum L(x*)Δx+
L(x *)Δx+ … L(x*)Δx =
Σ L(x*)Δx 12nii=1
of all such rectangles approximates the area of R.
In fact the mathematical definition of the area of R is the limit
n
Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞,
x
23. More on Areas
Δx
*
L(xi)
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
Δx
L(xn* )
xn*
a=x0 x b=x 1 x2 xi–1 xi
n
Σ L(xiL(x *)Δx 2*)Δx+ … L(xnThe Riemann sum *)Δx =
of all such rectangles approximates the area of R.
In fact the mathematical definition of the area of R is the limit
the area of R is A = ∫
b
L(x) dx.
x=a
R
L(x1*)Δx+
i=1
n
Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞, and
by the FTC
x
24. More on Areas
Δx
*
L(xi)
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
Δx
L(xn* )
xn*
a=x0 x b=x 1 x2 xi–1 xi
x
n
Σ L(xiL(x *)Δx 2*)Δx+ … L(xnThe Riemann sum *)Δx =
of all such rectangles approximates the area of R.
In fact the mathematical definition of the area of R is the limit
the area of R is A = ∫
b
L(x) dx.
x=a
R
L(x1*)Δx+
i=1
n
Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞, and
by the FTC
Theorem. The area of a 2D region is the definite integral of its
cross–section (length) function.
25. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
26. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
y = x2
y = –x2 + 2x
x
27. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area.
y = x2
y = –x2 + 2x
x
28. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
x
29. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x
30. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x x
31. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
x
32. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
x
33. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
L(x) =(–x2 + 2x) – x2
x x
L(x) =(–x2 + 2x) – x2
34. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
1
(–x2 + 2x) – x2 dx =
Hence the area is ∫
x=0
L(x) =(–x2 + 2x) – x2
x x
L(x) =(–x2 + 2x) – x2
35. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
1
(–x2 + 2x) – x2 dx = ∫
Hence the area is ∫
x=0
L(x) =(–x2 + 2x) – x2
x x
1
–2x2 + 2x dx
0
L(x) =(–x2 + 2x) – x2
36. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
1
(–x2 + 2x) – x2 dx = ∫
Hence the area is ∫
x=0
1
–2x2 + 2x dx
=
–2
3
1
x3 + x2 |
0
0
L(x) =(–x2 + 2x) – x2
x x
L(x) =(–x2 + 2x) – x2
37. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
1
(–x2 + 2x) – x2 dx = ∫
Hence the area is ∫
x=0
1
–2x2 + 2x dx
=
–2
3
1
=
x3 + x2 |
0
0
1
3
L(x) =(–x2 + 2x) – x2
x x
L(x) =(–x2 + 2x) – x2
38. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
39. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
f(x) = 2x – x3
g(x) = –x2
40. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts.
f(x) = 2x – x3
g(x) = –x2
41. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x).
f(x) = 2x – x3
g(x) = –x2
42. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
f(x) = 2x – x3
g(x) = –x2
43. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
44. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
45. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
46. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
47. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
48. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
Therefore the span for the 1st area is from x = –1 to x = 0
49. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
Therefore the span for the 1st area is from x = –1 to x = 0
and the span for the 2nd area is from x = 0 to x = 2.
50. 0
–x2 – (2x – x3) dx
∫
x= –1
More on Areas
f(x) = x – x3
2
2x – x3 – (–x2) dx
g(x) = –x2
Hence the total bounded area is
+ ∫
x= 0
–1 0 2
51. 0
–x2 – (2x – x3) dx
∫
x= –1
More on Areas
f(x) = x – x3
2
2x – x3 – (–x2) dx
g(x) = –x2
Hence the total bounded area is
+ ∫
x= 0
=
–1 0 2
0
x3 – x2 – 2x dx + ∫0
∫–1
2
– x3 + x2 + 2x dx
54. 0
–x2 – (2x – x3) dx
∫
x= –1
More on Areas
f(x) = x – x3
2
2x – x3 – (–x2) dx
2
g(x) = –x2
Hence the total bounded area is
+ ∫
x= 0
=
–1 0 2
0
x3 – x2 – 2x dx + ∫0
∫–1
2
– x3 + x2 + 2x dx
x4
4
= –
x3
3
– x2 |
0
–1
+
–x4
4
+ x3
3
+ x2 |
0
=
5
12
+
8
3
=
37
12
A type I region R is a region that
is bounded on top by a curve
y = f(x) and bounded below by
y = g(x) from x = a to x = b.
The above examples are regions
of type I.
55. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
56. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
y = b
y = a
x = g(y) x = f(y)
57. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
Its cross–sectional length is L(y) = f(y) – g(y).
y = b
y = a
x = g(y) x = f(y)
58. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
Its cross–sectional length is L(y) = f(y) – g(y).
y = b
y = a
x = g(y) x = f(y)
L(y) = f(y) – g(y)
59. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
Its cross–sectional length is L(y) = f(y) – g(y).
b
Therefore the area of R = ∫ f(y) – g(y) dy
y = b
y = a
y=a
x = g(y) x = f(y)
L(y) = f(y) – g(y)
60. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
61. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
y = x – 2
y = √x
62. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question.
y = x – 2
y = √x
63. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
64. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
65. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
66. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
67. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
68. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
4
69. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4.
4
70. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
4
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4. But the
lower boundary is not a single function.
71. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
4
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4. But the
lower boundary is not a single function.
72. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
4
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4. But the
lower boundary is not a single function. Therefore to find the
area of the region, we have to split it into two pieces,
I and II as shown.
73. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
I II
y = x – 2
y = √x
4
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4. But the
lower boundary is not a single function. Therefore to find the
area of the region, we have to split it into two pieces,
I and II as shown.
74. More on Areas
I II
y = x – 2
y = √x
2 4
The total area is (area I) + (area II) i.e.
75. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2 4
76. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
4
77. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
2
| +
0
4
78. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
4
79. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
=
2(2)3/2
3
+
4
80. More on Areas
I II
y = x – 2
y = √x
4
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
=
2(2)3/2
3
+
2(4)3/2
3
[( –
42
2
+ 8 ) –
2(2)3/2
–
3
22
2
( + 4 ) ]
81. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
=
2(2)3/2
3
+
2(4)3/2
3
[( –
42
2
+ 8 ) –
2(2)3/2
–
3
22
2
( + 4 ) ]
=
2(2)3/2
+
3
16
3
–
2(2)3/2
3
– 2
4
82. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
=
2(2)3/2
3
+
2(4)3/2
3
[( –
42
2
+ 8 ) –
2(2)3/2
–
3
22
2
( + 4 ) ]
=
2(2)3/2
+
3
16
3
–
2(2)3/2
3
– 2 =
10
3
4
83. More on Areas
y = x – 2
y = √x
However, we may view this as a type II region.
84. More on Areas
y = x – 2
y = √x
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2.
85. More on Areas
y = x – 2
y = √x
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
86. More on Areas
y = x – 2
y = √x
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2
87. More on Areas
y = x – 2
y = √x
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2
88. More on Areas
y = 2
y = √x
so x = y2
y = x – 2 so x = y + 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2
89. More on Areas
y = 2
y = √x
so x = y2
y = x – 2 so x = y + 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2
90. More on Areas
L(y) = y + 2 – y2
y = x – 2 so x = y + 2
y = √x
so x = y2
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.
91. More on Areas
L(y) = y + 2 – y2
y = x – 2 so x = y + 2
y = √x
so x = y2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.
Therefore the area is
2
∫ y + 2 – y2 dy
y = 0
y = 2
92. More on Areas
L(y) = y + 2 – y2
y = x – 2 so x = y + 2
y = √x
so x = y2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.
Therefore the area is
2
∫ y + 2 – y2 dy
y = 0
y2
2
= +
2y –
y3
|
2 3 0
y = 2
93. More on Areas
L(y) = y + 2 – y2
y = x – 2 so x = y + 2
y = √x
so x = y2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.
Therefore the area is
2
∫ y + 2 – y2 dy
y = 0
y2
2
= +
2y –
y3
|
2 3 0
= 6 –
8
3
=
10
3
y = 2