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IOSR Journal of Mathematics(IOSR-JM) is a double blind peer reviewed International Journal that provides rapid publication (within a month) of articles in all areas of mathemetics and its applications. The journal welcomes publications of high quality papers on theoretical developments and practical applications in mathematics. Original research papers, state-of-the-art reviews, and high quality technical notes are invited for publications.
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At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
the slides are aimed to give a brief introductory base to Neural Networks and its architectures. it covers logistic regression, shallow neural networks and deep neural networks. the slides were presented in Deep Learning IndabaX Sudan.
The second Fundamental Theorem of Calculus makes calculating definite integrals a problem of antidifferentiation!
(the slideshow has extra examples based on what happened in class)
Lesson 24: Areas, Distances, the Integral (Section 021 slides)Matthew Leingang
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The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
Estimation of Distribution Algorithms (EDAs) constitute a powerful evolutionary algorithm for solving continuous and combinatorial optimization problems. Based on machine learning techniques, at each generation, EDAs estimate a joint probability distribution associated with the set of most promising individuals, trying to explicitly express the interrelations between the different variables of the problem. Based on this general framework, EDAs have proved to be very competitive for solving combinatorial and continuous optimization problems. In this talk, we propose developing EDAs by introducing probability models defined exclusively on the space of feasible solutions. In this sense, we give a first approach by taking the Graph Partitioning Problem (GPP) as a case of study, and present a probabilistic model defined exclusively on the feasible region of solutions: a square lattice probability model.
Similarity Measure Using Interval Valued Vague Sets in Multiple Criteria Deci...iosrjce
IOSR Journal of Mathematics(IOSR-JM) is a double blind peer reviewed International Journal that provides rapid publication (within a month) of articles in all areas of mathemetics and its applications. The journal welcomes publications of high quality papers on theoretical developments and practical applications in mathematics. Original research papers, state-of-the-art reviews, and high quality technical notes are invited for publications.
Mixed Spectra for Stable Signals from Discrete Observationssipij
This paper concerns the continuous-time stable alpha symmetric processes which are inivitable in the modeling of certain signals with indefinitely increasing variance. Particularly the case where the spectral measurement is mixed: sum of a continuous measurement and a discrete measurement. Our goal is to estimate the spectral density of the continuous part by observing the signal in a discrete way. For that, we propose a method which consists in sampling the signal at periodic instants. We use Jackson's polynomial kernel to build a periodogram which we then smooth by two spectral windows taking into account the width of the interval where the spectral density is non-zero. Thus, we bypass the phenomenon of aliasing often encountered in the case of estimation from discrete observations of a continuous time process.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
the slides are aimed to give a brief introductory base to Neural Networks and its architectures. it covers logistic regression, shallow neural networks and deep neural networks. the slides were presented in Deep Learning IndabaX Sudan.
The second Fundamental Theorem of Calculus makes calculating definite integrals a problem of antidifferentiation!
(the slideshow has extra examples based on what happened in class)
Lesson 24: Areas, Distances, the Integral (Section 021 slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 15: Exponential Growth and Decay (Section 021 slides)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Lesson 12: Linear Approximations and Differentials (slides)Matthew Leingang
The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
Lesson 24: Areas, Distances, the Integral (Section 041 slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
JEE Mathematics/ Lakshmikanta Satapathy/ Set theory part 3/ JEE question on Power set of the cartesian product of two sets solved with the related concepts
JEE Mathematics/ Lakshmikanta Satapathy/ Permutation and Combination QA part 2/ JEE question on four digit numbers permuted from five given digits solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Wave Motion QA part 3/ JEE question on fundamental frequencies of Open and Closed pipes solved with the related concepts
We define the definite integral as a limit of Riemann sums, compute some approximations, then investigate the basic additive and comparative properties
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
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back to students electronically. AMC provides an email gateway, but I have written programs to return graded papers via the DAV protocol to student’s dropboxes on our (Sakai) learning management systems. I will also show how graded papers can be archived, with appropriate metadata tags, into an Evernote notebook.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Lesson 24: Areas and Distances, The Definite Integral (handout)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 24: Areas and Distances, The Definite Integral (slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
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The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
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Topics covered:
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UI automation Sample
Desktop automation flow
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Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
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1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
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Execution from the test manager
Orchestrator execution result
Defect reporting
SAP heatmap example with demo
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Send an interactive Slack channel message (using buttons)
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In a second workflow supporting the same use case, you’ll see:
Your campaign sent to target colleagues for approval
If the “Approve” button is clicked, a Jira/Zendesk ticket is created for the marketing design team
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See how to accelerate model training and optimize model performance with active learning
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Get an exclusive demo of the new family of UiPath LLMs – GenAI models specialized for processing different types of documents and messages
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All of this illustrated with link prediction over knowledge graphs, but the argument is general.
Search and Society: Reimagining Information Access for Radical FuturesBhaskar Mitra
The field of Information retrieval (IR) is currently undergoing a transformative shift, at least partly due to the emerging applications of generative AI to information access. In this talk, we will deliberate on the sociotechnical implications of generative AI for information access. We will argue that there is both a critical necessity and an exciting opportunity for the IR community to re-center our research agendas on societal needs while dismantling the artificial separation between the work on fairness, accountability, transparency, and ethics in IR and the rest of IR research. Instead of adopting a reactionary strategy of trying to mitigate potential social harms from emerging technologies, the community should aim to proactively set the research agenda for the kinds of systems we should build inspired by diverse explicitly stated sociotechnical imaginaries. The sociotechnical imaginaries that underpin the design and development of information access technologies needs to be explicitly articulated, and we need to develop theories of change in context of these diverse perspectives. Our guiding future imaginaries must be informed by other academic fields, such as democratic theory and critical theory, and should be co-developed with social science scholars, legal scholars, civil rights and social justice activists, and artists, among others.
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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.
1. .
Section 5.3
Evaluating Definite Integrals
V63.0121.041, Calculus I
New York University
December 6, 2010
Announcements
Today: Section 5.3
Wednesday: Section 5.4
Monday, December 13: Section 5.5
”Monday,” December 15: Review and Movie Day!
Monday, December 20, 12:00–1:50pm: Final Exam
. . . . . .
2. Announcements
Today: Section 5.3
Wednesday: Section 5.4
Monday, December 13:
Section 5.5
”Monday,” December 15:
Review and Movie Day!
Monday, December 20,
12:00–1:50pm: Final Exam
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 2 / 41
3. Objectives
Use the Evaluation
Theorem to evaluate
definite integrals.
Write antiderivatives as
indefinite integrals.
Interpret definite integrals
as “net change” of a
function over an interval.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 3 / 41
4. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 4 / 41
5. The definite integral as a limit
Definition
If f is a function defined on [a, b], the definite integral of f from a to b
is the number ∫ b ∑n
f(x) dx = lim f(ci ) ∆x
a n→∞
i=1
b−a
where ∆x = , and for each i, xi = a + i∆x, and ci is a point in
n
[xi−1 , xi ].
Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite integral
∫ b
f(x) dx exists and is the same for any choice of ci .
a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 5 / 41
6. Notation/Terminology
∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integration (a is the lower limit and b the
upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of computing an integral is called integration
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 6 / 41
7. Example
∫ 1
4
Estimate dx using the midpoint rule and four divisions.
0 1 + x2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
8. Example
∫ 1
4
Estimate dx using the midpoint rule and four divisions.
0 1 + x2
Solution
Dividing up [0, 1] into 4 pieces gives
1 2 3 4
x0 = 0, x1 = , x2 = , x3 = , x4 =
4 4 4 4
So the midpoint rule gives
( )
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
11. Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant. Then
∫ b
1. c dx = c(b − a)
a
∫ b ∫ b ∫ b
2. [f(x) + g(x)] dx = f(x) dx + g(x) dx.
a a a
∫ b ∫ b
3. cf(x) dx = c f(x) dx.
a a
∫ b ∫ b ∫ b
4. [f(x) − g(x)] dx = f(x) dx − g(x) dx.
a a a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 8 / 41
12. More Properties of the Integral
Conventions: ∫ ∫
a b
f(x) dx = − f(x) dx
b a
∫ a
f(x) dx = 0
a
This allows us to have
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 9 / 41
13. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
.
a b c x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
14. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b
f(x) dx
a
.
a b c x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
15. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b ∫ c
f(x) dx f(x) dx
a b
.
a b c x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
16. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b ∫ c ∫ c
f(x) dxf(x) dxf(x) dx
a a b
.
a b c x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
17. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
.
a c x
b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
18. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ b
f(x) dx
a
.
a c x
b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
19. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ c
f(x) dx =
b∫
b
− f(x) dx
c
.
a c x
b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
20. Illustrating Property 5
Theorem
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
y
∫ c ∫ c
f(x) dx f(x) dx =
a b∫
b
− f(x) dx
c
.
a c x
b
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
21. Definite Integrals We Know So Far
If the integral computes an
area and we know the
area, we can use that. For
instance,
y
∫ 1√
π
1 − x2 dx =
0 4
By brute force we
.
computed x
∫ 1 ∫ 1
1 1
x2 dx = x3 dx =
0 3 0 4
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 11 / 41
22. Comparison Properties of the Integral
Theorem
Let f and g be integrable functions on [a, b].
∫ b
6. If f(x) ≥ 0 for all x in [a, b], then f(x) dx ≥ 0
a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
23. Comparison Properties of the Integral
Theorem
Let f and g be integrable functions on [a, b].
∫ b
6. If f(x) ≥ 0 for all x in [a, b], then f(x) dx ≥ 0
a
7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b ∫ b
f(x) dx ≥ g(x) dx
a a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
24. Comparison Properties of the Integral
Theorem
Let f and g be integrable functions on [a, b].
∫ b
6. If f(x) ≥ 0 for all x in [a, b], then f(x) dx ≥ 0
a
7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b ∫ b
f(x) dx ≥ g(x) dx
a a
8. If m ≤ f(x) ≤ M for all x in [a, b], then
∫ b
m(b − a) ≤ f(x) dx ≤ M(b − a)
a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
25. Estimating an integral with inequalities
Example
∫ 2
1
Estimate dx using Property 8.
1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
26. Estimating an integral with inequalities
Example
∫ 2
1
Estimate dx using Property 8.
1 x
Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
we have ∫ 2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or ∫ 2
1 1
≤ dx ≤ 1
2 1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
27. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 14 / 41
28. Socratic proof
The definite integral of
velocity measures
displacement (net
distance)
The derivative of
displacement is velocity
So we can compute
displacement with the
definite integral or the
antiderivative of velocity
But any function can be a
velocity function, so . . .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 15 / 41
29. Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′ for another function F, then
∫ b
f(x) dx = F(b) − F(a).
a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
30. Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′ for another function F, then
∫ b
f(x) dx = F(b) − F(a).
a
Note
In Section 5.3, this theorem is called “The Evaluation Theorem”.
Nobody else in the world calls it that.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
31. Proving the Second FTC
b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual.
n
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
32. Proving the Second FTC
b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual.
n
For each i, F is continuous on [xi−1 , xi ] and differentiable on
(xi−1 , xi ). So there is a point ci in (xi−1 , xi ) with
F(xi ) − F(xi−1 )
= F′ (ci ) = f(ci )
xi − xi−1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
33. Proving the Second FTC
b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual.
n
For each i, F is continuous on [xi−1 , xi ] and differentiable on
(xi−1 , xi ). So there is a point ci in (xi−1 , xi ) with
F(xi ) − F(xi−1 )
= F′ (ci ) = f(ci )
xi − xi−1
Or
f(ci )∆x = F(xi ) − F(xi−1 )
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
34. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
35. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
36. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
37. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
38. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
39. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
40. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
41. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
42. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
43. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
44. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
45. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
46. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
47. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
48. Proof continued
We have for each i
f(ci )∆x = F(xi ) − F(xi−1 )
Form the Riemann Sum:
∑
n ∑
n
Sn = f(ci )∆x = (F(xi ) − F(xi−1 ))
i=1 i=1
= (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · ·
· · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 ))
= F(xn ) − F(x0 ) = F(b) − F(a)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
49. Proof Completed
We have shown for each n,
Sn = F(b) − F(a)
Which does not depend on n.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 19 / 41
50. Proof Completed
We have shown for each n,
Sn = F(b) − F(a)
Which does not depend on n.
So in the limit
∫ b
f(x) dx = lim Sn = lim (F(b) − F(a)) = F(b) − F(a)
a n→∞ n→∞
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 19 / 41
51. Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
52. Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
∫ 1 1
x4 1
A= x3 dx = =
0 4 0 4 .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
53. Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
∫ 1 1
x4 1
A= x3 dx = =
0 4 0 4 .
Here we use the notation F(x)|b or [F(x)]b to mean F(b) − F(a).
a a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
54. Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
55. Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
1
.
−1 1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
56. Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
1
.
−1 1
Solution
∫ 1 [ ]1 ( [ )]
x3 1 1 4
A=2− x dx = 2 − 2
=2− − − =
−1 3 −1 3 3 3
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
57. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 22 / 41
59. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
Solution
∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
60. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
Solution
∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
61. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
Solution
∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
62. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
Solution
∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)
(π )
=4 −0
4
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
63. Computing an integral we estimated before
Example
∫ 1
4
Evaluate the integral dx.
0 1 + x2
Solution
∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)
(π )
=4 −0 =π
4
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
64. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 25 / 41
65. Estimating an integral with inequalities
Example
∫ 2
1
Estimate dx using Property 8.
1 x
Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
we have ∫ 2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or ∫ 2
1 1
≤ dx ≤ 1
2 1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 26 / 41
66. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solution
∫ 2
1
dx
1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
67. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solution
∫ 2
1
dx = ln x|2
1
1 x
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
68. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solution
∫ 2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
69. Computing an integral we estimated before
Example
∫ 2
1
Evaluate dx.
1 x
Solution
∫ 2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
= ln 2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
70. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 28 / 41
71. The Integral as Total Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramifications:
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
72. The Integral as Total Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramifications:
Theorem
If v(t) represents the velocity of a particle moving rectilinearly, then
∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
73. The Integral as Total Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramifications:
Theorem
If MC(x) represents the marginal cost of making x units of a product,
then ∫ x
C(x) = C(0) + MC(q) dq.
0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
74. The Integral as Total Change
Another way to state this theorem is:
∫ b
F′ (x) dx = F(b) − F(a),
a
or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramifications:
Theorem
If ρ(x) represents the density of a thin rod at a distance of x from its
end, then the mass of the rod up to x is
∫ x
m(x) = ρ(s) ds.
0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
75. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 30 / 41
76. A new notation for antiderivatives
To emphasize the relationship between antidifferentiation and
integration, we use the indefinite integral notation
∫
f(x) dx
for any function whose derivative is f(x).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 31 / 41
77. A new notation for antiderivatives
To emphasize the relationship between antidifferentiation and
integration, we use the indefinite integral notation
∫
f(x) dx
for any function whose derivative is f(x). Thus
∫
x2 dx = 1 x3 + C.
3
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 31 / 41
78. My first table of integrals
.
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫ ∫
xn+1
n
x dx = + C (n ̸= −1) cf(x) dx = c f(x) dx
n+1 ∫
∫
1
ex dx = ex + C dx = ln |x| + C
x
∫ ∫
ax
sin x dx = − cos x + C ax dx = +C
ln a
∫ ∫
cos x dx = sin x + C csc2 x dx = − cot x + C
∫ ∫
sec2 x dx = tan x + C csc x cot x dx = − csc x + C
∫ ∫
1
sec x tan x dx = sec x + C √ dx = arcsin x + C
∫ 1 − x2
1
dx = arctan x + C
1 + x2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 32 / 41
79. Outline
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 33 / 41
80. Computing Area with integrals
Example
Find the area of the region bounded by the lines x = 1, x = 4, the
x-axis, and the curve y = ex .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 34 / 41
81. Computing Area with integrals
Example
Find the area of the region bounded by the lines x = 1, x = 4, the
x-axis, and the curve y = ex .
Solution
The answer is ∫ 4
4
ex dx = ex |1 = e4 − e.
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 34 / 41
82. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
83. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.
.
x
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
84. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.
Instead compute the area as
∫ π/2
π
− sin y dy .
2 0
x
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
85. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.
Instead compute the area as
∫ π/2
π π π/2
− sin y dy = −[− cos x]0 .
2 0 2
x
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
86. Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.
Solution
∫ 1
The answer is arcsin x dx, but we y
0
do not know an antiderivative for π/2
arcsin.
Instead compute the area as
∫ π/2
π π π/2 π
− sin y dy = −[− cos x]0 = −1 .
2 0 2 2
x
1
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
87. Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and
the vertical lines x = 0 and x = 3.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 36 / 41
88. Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and
the vertical lines x = 0 and x = 3.
Solution
∫ 3
Consider (x − 1)(x − 2) dx.
0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 36 / 41
89. Graph
y
. x
1 2 3
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 37 / 41
90. Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and
the vertical lines x = 0 and x = 3.
Solution
∫ 3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
0
and (2, 3], and negative on (1, 2).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 38 / 41
91. Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and
the vertical lines x = 0 and x = 3.
Solution
∫ 3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
0
and (2, 3], and negative on (1, 2). If we want the area of the region, we
have to do
∫ 1 ∫ 2 ∫ 3
A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx
0 1 2
[ ]1 [ ]2 [ ]3
= − 1 3
3x
3 2
2x + 2x − 1 x3 − 3 x2 + 2x + 1 x3 − 3 x2 + 2x
3 2 3 2
( ) 0 1 2
5 1 5 11
= − − + = .
6 6 6 6
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 38 / 41
92. Interpretation of “negative area" in motion
There is an analog in rectlinear motion:
∫ t1
v(t) dt is net distance traveled.
t0
∫ t1
|v(t)| dt is total distance traveled.
t0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 39 / 41
93. What about the constant?
It seems we forgot about the +C when we say for instance
∫ 1 1
x4 1 1
3
x dx = = −0=
0 4 0 4 4
But notice
[ 4 ]1 ( )
x 1 1 1
+C = + C − (0 + C) = + C − C =
4 0 4 4 4
no matter what C is.
So in antidifferentiation for definite integrals, the constant is
immaterial.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 40 / 41
94. Summary
The second Fundamental Theorem of Calculus:
∫ b
f(x) dx = F(b) − F(a)
a
where F′ = f.
Definite integrals represent net change of a function over an
interval. ∫
We write antiderivatives as indefinite integrals f(x) dx
. . . . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 41 / 41