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.
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.
The first report of Machine Learning Seminar organized by Computational Linguistics Laboratory at Kazan Federal University. See http://cll.niimm.ksu.ru/cms/lang/en_US/main/seminars/mlseminar
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.
Lesson 12: Linear Approximation (Section 41 handout)Matthew Leingang
The line tangent to a curve, which 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.
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.
3D gravity inversion by planting anomalous densitiesLeonardo Uieda
Paper presented at the 2011 SBGf International Congress in Rio de Janeiro, Brazil.
Abstract:
This paper presents a novel gravity inversion method for estimating a 3D density-contrast distribution defined on a grid of prisms. Our method consists of an iterative algorithm that does not require the solution of a large equation system. Instead, the solution
grows systematically around user-specified prismatic
elements called “seeds”. Each seed can have a different density contrast, allowing the interpretation of multiple bodies with different density contrasts and interfering gravitational effects. The compactness of the solution around the seeds is imposed by
means of a regularizing function. The solution
grows by the accretion of neighboring prisms of the
current solution. The prisms for the accretion are chosen by systematically searching the set of current neighboring prisms. Therefore, this approach allows that the columns of the Jacobian matrix be calculated on demand. This is a known technique from computer science called “lazy evaluation”, which greatly reduces the demand of computer memory and processing time. Test on synthetic data and on real data collected over the ultramafic Cana Brava complex, central Brazil, confirmed the ability of our method in detecting sharp
and compact bodies.
The first report of Machine Learning Seminar organized by Computational Linguistics Laboratory at Kazan Federal University. See http://cll.niimm.ksu.ru/cms/lang/en_US/main/seminars/mlseminar
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.
Lesson 12: Linear Approximation (Section 41 handout)Matthew Leingang
The line tangent to a curve, which 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.
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.
3D gravity inversion by planting anomalous densitiesLeonardo Uieda
Paper presented at the 2011 SBGf International Congress in Rio de Janeiro, Brazil.
Abstract:
This paper presents a novel gravity inversion method for estimating a 3D density-contrast distribution defined on a grid of prisms. Our method consists of an iterative algorithm that does not require the solution of a large equation system. Instead, the solution
grows systematically around user-specified prismatic
elements called “seeds”. Each seed can have a different density contrast, allowing the interpretation of multiple bodies with different density contrasts and interfering gravitational effects. The compactness of the solution around the seeds is imposed by
means of a regularizing function. The solution
grows by the accretion of neighboring prisms of the
current solution. The prisms for the accretion are chosen by systematically searching the set of current neighboring prisms. Therefore, this approach allows that the columns of the Jacobian matrix be calculated on demand. This is a known technique from computer science called “lazy evaluation”, which greatly reduces the demand of computer memory and processing time. Test on synthetic data and on real data collected over the ultramafic Cana Brava complex, central Brazil, confirmed the ability of our method in detecting sharp
and compact bodies.
Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.
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 derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
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.
Similar to Lesson 27: Integration by Substitution (Section 021 handout) (20)
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
Auto-multiple-choice (AMC) is an open-source optical mark recognition software package built with Perl, LaTeX, XML, and sqlite. I use it for all my in-class quizzes and exams. Unique papers are created for each student, fixed-response items are scored automatically, and free-response problems, after manual scoring, have marks recorded in the same process. In the first part of the talk I will discuss AMC’s many features and why I feel it’s ideal for a mathematics course. My contributions to the AMC workflow include some scripts designed to automate the process of returning scored papers
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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.
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.
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.
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...James Anderson
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https://alandix.com/academic/papers/synergy2024-epistemic/
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Lesson 27: Integration by Substitution (Section 021 handout)
1. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
Notes
Section 5.5
Integration by Substitution
V63.0121.021, Calculus I
New York University
December 14, 2010
Announcements
”Wednesday”, December 15: Review, Movie
Monday, December 20, 12:00pm–1:50pm: Final Exam
Announcements
Notes
”Wednesday”, December 15:
Review, Movie
Monday, December 20,
12:00pm–1:50pm: Final
Exam
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 2 / 37
Objectives
Notes
Given an integral and a
substitution, transform the
integral into an equivalent
one using a substitution
Evaluate indefinite integrals
using the method of
substitution.
Evaluate definite integrals
using the method of
substitution.
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 4 / 37
1
2. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
Outline
Notes
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 5 / 37
Differentiation and Integration as reverse processes
Notes
Theorem (The Fundamental Theorem of Calculus)
1. Let f be continuous on [a, b]. Then
x
d
f (t) dt = f (x)
dx a
2. Let f be continuous on [a, b] and f = F for some other function F .
Then
b
f (x) dx = F (b) − F (a).
a
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 6 / 37
Techniques of antidifferentiation?
Notes
So far we know only a few rules for antidifferentiation. Some are general,
like
[f (x) + g (x)] dx = f (x) dx + g (x) dx
Some are pretty particular, like
1
√ dx = arcsec x + C .
x x2 − 1
What are we supposed to do with that?
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 7 / 37
2
3. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
No straightforward system of antidifferentiation
Notes
So far we don’t have any way to find
2x
√ dx
x2 + 1
or
tan x dx.
Luckily, we can be smart and use the “anti” version of one of the most
important rules of differentiation: the chain rule.
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 8 / 37
Outline
Notes
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 9 / 37
Substitution for Indefinite Integrals
Notes
Example
Find
x
√ dx.
x2 + 1
Solution
Stare at this long enough and you notice the the integrand is the
derivative of the expression 1 + x 2 .
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 10 / 37
3
4. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
Say what?
Notes
Solution (More slowly, now)
Let g (x) = x 2 + 1. Then g (x) = 2x and so
d 1 x
g (x) = g (x) = √
dx 2 g (x) x2 + 1
Thus
x d
√ dx = g (x) dx
x2 + 1 dx
= g (x) + C = 1 + x2 + C .
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 11 / 37
Leibnizian notation FTW
Notes
Solution (Same technique, new notation)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand
becomes completely transformed into
1
√
x dx 2 du
√
1
√ du
= =
x2 + 1 u 2 u
1 −1/2
= 2u du
√
= u+C = 1 + x2 + C .
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 12 / 37
Useful but unsavory variation
Notes
Solution (Same technique, new notation, more idiot-proof)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
du
dx =
2x
So the integrand becomes completely transformed into
x x du 1
√ dx = √ · = √ du
x2 + 1 u 2x 2 u
1 −1/2
= 2u du
√
= u+C = 1 + x2 + C .
Mathematicians have serious issues with mixing the x and u like this.
However, I can’t deny that it works.
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 13 / 37
4
5. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
Theorem of the Day
Notes
Theorem (The Substitution Rule)
If u = g (x) is a differentiable function whose range is an interval I and f
is continuous on I , then
f (g (x))g (x) dx = f (u) du
That is, if F is an antiderivative for f , then
f (g (x))g (x) dx = F (g (x))
In Leibniz notation:
du
f (u) dx = f (u) du
dx
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 14 / 37
A polynomial example
Notes
Example
Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx.
Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So
(x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du
1 1
= u 4 = (x 2 + 3)4
2 2
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 15 / 37
A polynomial example, by brute force
Notes
Compare this to multiplying it out:
(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx
= 4x 7 + 36x 5 + 108x 3 + 108x dx
1
= x 8 + 6x 6 + 27x 4 + 54x 2
2
Which would you rather do?
It’s a wash for low powers
But for higher powers, it’s much easier to do substitution.
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 16 / 37
5
6. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
Compare
Notes
We have the substitution method, which, when multiplied out, gives
1
(x 2 + 3)3 4x dx = (x 2 + 3)4
2
1 8
= x + 12x 6 + 54x 4 + 108x 2 + 81
2
1 81
= x 8 + 6x 6 + 27x 4 + 54x 2 +
2 2
and the brute force method
1
(x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2
2
Is there a difference? Is this a problem?
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 17 / 37
A slick example
Notes
Example
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solution
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 18 / 37
Outline
Notes
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 21 / 37
6
7. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
Substitution for Definite Integrals
Notes
Theorem (The Substitution Rule for Definite Integrals)
If g is continuous and f is continuous on the range of u = g (x), then
b g (b)
f (g (x))g (x) dx = f (u) du.
a g (a)
Why the change in the limits?
The integral on the left happens in “x-land”
The integral on the right happens in “u-land”, so the limits need to
be u-values
To get from x to u, apply g
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 22 / 37
Example
π
Notes
Compute cos2 x sin x dx.
0
Solution (Slow Way)
First compute the indefinite integral cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and
cos2 x sin x dx = − u 2 du
= − 3 u 3 + C = − 1 cos3 x + C .
1
3
Therefore
π π
1 1 2
cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = .
0 3 0 3 3
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 23 / 37
Definite-ly Quicker
Notes
Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
π −1 1
cos2 x sin x dx = −u 2 du = u 2 du
0 1 −1
1
1 3 1 2
= u = 1 − (−1) =
3 −1 3 3
The advantage to the “fast way” is that you completely transform the
integral into something simpler and don’t have to go back to the
original variable (x).
But the slow way is just as reliable.
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 24 / 37
7
8. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
An exponential example
Notes
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 25 / 37
Another way to skin that cat
Notes
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 30 / 37
A third skinned cat
Notes
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 31 / 37
8
9. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
A Trigonometric Example
Notes
Example
Find
3π/2
θ θ
cot5 sec2 dθ.
π 6 6
Before we dive in, think about:
What “easy” substitutions might help?
Which of the trig functions suggests a substitution?
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 32 / 37
Solution
Notes
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 33 / 37
Graphs
3π/2
θ θ π/4 Notes
cot5 sec2 dθ 6 cot5 ϕ sec2 ϕ dϕ
π 6 6 π/6
y y
θ ϕ
3π π ππ
2 64
The areas of these two regions are the same.
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 35 / 37
9
10. V63.0121.021, Calculus I Section 5.5 : Integration by Substitution December 14, 2010
Graphs
π/4 1
−5 Notes
6 cot5 ϕ sec2 ϕ dϕ √ 6u du
π/6 1/ 3
y y
ϕ u
ππ 1 1
64 √
3
The areas of these two regions are the same.
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 36 / 37
Summary
Notes
If F is an antiderivative for f , then:
f (g (x))g (x) dx = F (g (x))
If F is an antiderivative for f , which is continuous on the range of g ,
then:
b g (b)
f (g (x))g (x) dx = f (u) du = F (g (b)) − F (g (a))
a g (a)
Antidifferentiation in general and substitution in particular is a
“nonlinear” problem that needs practice, intuition, and perserverance
The whole antidifferentiation story is in Chapter 6
V63.0121.021, Calculus I (NYU) Section 5.5 Integration by Substitution December 14, 2010 37 / 37
Notes
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