Change of Variable and Integration by Substitution
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We use the Chain rule to derive the Integration by Substitution formula, and then investigate the remarkable Change of Variable formula that shows that the notion of signed area, even if defined originally in terms of a parametrized curve, is actually independent of that parametrization, in the sense that if we compose with a change of variable, the signed area stays unchanged. Through some examples we will see that this leads us to have a more flexible idea of what a "curve segment" actually means. And we provide a purely algebraic proof, based on Faulhaber's algebraic Derivative D and Integral S, that completely avoids the usual physical or geometric arguments involving limits, and transformations of small or infinitesimal quantities.
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[Music]
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hello everyone I'm Norval burger and
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welcome to the algebraic calculus in
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today's lecture we're going to be
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talking about change of variables and
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integration by substitution so this is a
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standard and important technique for
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helping us to augment our ability to
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integrate and also the actual change of
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variables formula is a really a
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remarkable aspect of the theory of
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integration that we've developed going
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back has to think essentially to live
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notes so recall that for the file hopper
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derivative capital D we have our
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familiar product rule now there's a sort
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of special consequence of this called
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the product power rule that the
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derivative of P to the N is n times P to
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the N minus 1 times D of P for a general
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Pauline umber P and I also remind you of
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the chain rule that if we define the
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composition of two polymers say P circle
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Q of alpha is P of Q of alpha then the
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derivative of the composition P circle Q
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is the composition of the derivative of
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P with Q times the derivative of Q and I
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remind you that this one here follows
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from this one here by induction okay
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once you know how the derivative of the
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product of two things works then you can
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figure out what the derivative of a
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product of n things works by it sort of
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applying it over and over again and then
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to go from here to here it's a question
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of looking at P and sort of separating
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it out P is a polynomial of terms each
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of a certain power and basically
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applying this formula to each one of
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those things okay so that's sort of
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logical structure of these things and
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today we're going to be looking at this
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chain rule and what happens we're sort
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of when we integrate it well we can
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extract from that
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I remind you that when we derive the
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integration by parts formula we combine
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the product rule with the derivative
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integral composition theorem we're gonna
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do the same kind of thing today but
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we're going to combine this thing now
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with a chain rule so this derivative
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integral composition theorem tells us
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what happens if we first integrate and
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then differentiate first of all we get
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back exactly what we started that's the
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identity operation however if we do the
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derivative first and then we integrate
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then we almost get back what we started
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it's the identity minus I 0 that means
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we have to peel off the constant term of
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the polynomial that we're starting with
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that's the actual top element just is
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replacement 0 ok so if we combine this
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with the chain rule we get the algebraic
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integration by substitution formula here
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it is they file harbor integral capital
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S from T to U of this expression the
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derivative of P composed with Q times
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the derivative of Q that equals the
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composition P circle Q evaluated from T
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to you and why is that well because the
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chain rule tells us that the derivative
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of this is exactly what's inside here so
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if we take the integral of this we're
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essentially getting this thing back
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minus the identity minus the zero of
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part of it and we get rid of that zeroth
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part by evaluating the integral between
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two different values T and u the value
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at u minus the value of T then the the
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constant term is disappearing so that's
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why we're just getting this P circle Q
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from T to you that's the integral of
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this composition okay so for an example
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let's calculate the integral from 1 to 3
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of alpha times 1 plus alpha squared to
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the fifth now the tricky is to kind of
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arrange this in your mind so it looks
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like the composition of something with a
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derivative of the thing that we're
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composing
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with so what we can do here in our minds
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is to set P equals alpha to the six over
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six and Q equals one plus alpha squared
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so now the derivative of Q is 2 alpha
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that's more or less that term there and
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the derivative of P will be alpha to the
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fifth and what's appearing here is that
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derivative composed with Q so it's 1
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plus alpha squared to the fifth okay
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so except for a constant term here we're
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almost exactly in the situation of
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having identified how the terms here
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correspond to the terms here by a
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suitable choice of P and Q so this thing
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is in fact we're just sort of
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rearranging it now to make it closer to
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2 what we want so we'll rewrite it as
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integral from 1 to 3 1 plus alpha
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squared to the fifth times 2 alpha now
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and we'll multiply by 1/2 here to
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compensate that to that we've introduced
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there so now we really have this thing
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here is the derivative of P composed
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with Q that's exactly what that is and
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this term here is the derivative of Q so
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the algebraic integration by
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substitution formula tells us that this
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is going to be or the half there and
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then this whole thing will be P Circle Q
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so the P composed with Q so that's that
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inside there that's P Circle Q we're
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taking Q to the sixth power and dividing
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by 6 okay so that's relatively easy we
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just plug in 3 we get 1 plus 9 that's 10
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to the 6th minus 1 plus 1 that's 2 to
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the sixth the 6 comes out front with a
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112 there that's an example of an
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integration by substitution and again
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this is a purely algebraic thing it's
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just at the level of foul Harbor
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integration Feilhaber derivatives just
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purely in terms of polynomials
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right now we come to a really remarkable
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formula closely related to the
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integration by substitution formula this
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is the change of variable formula that
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allows us to re parameterize a curve and
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we still get the same signed area so
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it's telling us that somehow the signed
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area really is associated to the
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geometrical curve whatever that is
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rather than dependent on the particular
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parameter ization that we've chosen but
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we'll see in the examples that it's a
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little bit more subtle than that
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anyway here is the formula for poly
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numbers P Q and R if we take the signed
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area from T to U of the curve defined by
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P circle R and Q circle R that's equal
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to the signed area from R of T to R of U
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of P Q so the original curve may be this
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one here
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defined by P and Q and we're going from
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R of T to R of U that's the same as this
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more complicated composition but
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evaluated at the more simple bounds of T
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and u so this remarkable formula
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suggests that the repr amat rising a
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curve does not actually change signed
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areas and it's essentially an
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observation a key observation of live
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notes okay so now I want to illustrate
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the change of variable formula with a
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series of examples starting with some
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simple ones and then working our way up
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and as we go we're gonna see that what
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exactly a curve is is going to be called
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into question a little bit okay so we're
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gonna have to be flexible or about our
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understanding of the geometry that goes
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with the algebra so let's start with a
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simple case essentially parabola where P
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is alpha and Q is alpha squared okay so
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that's the curve usually called y equals
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x squared and we're going to compose
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these with 1 plus 3 alpha call that R so
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now we have 1 plus 3 alpha 1 plus 3
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alpha squared and we're going to agree
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that we're going
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from zero to one when alpha 0 we have
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0.11 when alphas 1 we have the point 4 4
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squared up there for 16 so we're talking
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about this portion of the curve so let's
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calculate its signed area signed area
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from TD U of P circle our Q circle R so
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the compositions are 1 plus 3 alpha and
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1 plus 3 alpha squared
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okay we're composing 1 plus 3 alpha with
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both of these okay so this is a signed
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area we know how to compute that we can
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use the fundamental theorem of algebra
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calculus saying that we can calculate by
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1/2 of the foul harbor integral from 0
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to 1 of the first term times the
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derivative of the second minus the
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second times the derivative of the first
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ok the derivative of 1 plus 3 alpha
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squared is well there'll be 2 times 1
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plus 3 alpha with another 3 so there
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will be then 1 plus 3 alpha squared with
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a 2 and also a factor of 3 appearing and
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over here the derivative of this is the
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is just 3 so we get 3 times 1 plus 3
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alpha squared and these combine to give
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us 3 times 1 plus 3 alpha squared okay
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this is easy to integrate the integral
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is 1 plus 3 alpha all cubed over 3 you
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can check that the derivative of this is
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what's inside here and we're going from
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a 0 to 1 so we substitute 1 at which
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point we get 4 cubed substitute 0 we get
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1 cubed this thing comes out denominator
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6 there so all together we get 63 over 6
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for the signed area of this thing
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computed from this point of view when
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we're thinking about it as being
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expressed this way with the
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parameterization 1 plus 3 alpha 1 plus 3
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alpha squared from 0 to 1 now on the
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other hand if we just sort of stick with
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the curve peak
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you alpha alpha squared now we have to
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go from R of T to R of you R of T R zero
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is one R of U is four so it's the signed
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area from one to four of alpha alpha
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squared so it's the same curve sort of
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segment geometrically but it's
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parametrized in a different way and this
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integral is very easy to do it's just
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one six because the fundamental formula
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to minus 1 over 2 times 2 plus 1 that's
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1 6 and then there's a power of 3 so
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it's 4 cubed minus 1 cubed and we get
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the same answer so the long and short of
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it is that from going from here to here
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we've Reaper amortized the curve the
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calculation of the integral is then
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different but you end up with the same
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answer
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so the signed area is at least in this
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case independent of the parameterization
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at least we've been able to change from
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this parameterization to this one and
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this signed area has not changed now
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let's have a look at a little bit more
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sophisticated example of this change of
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variables formula so we've got a curve
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that looks like this now given by p
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equals alpha cubed and Q equals 4 plus
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alpha squared and the change of variable
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is given by R equals 1 minus 2 alpha
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plus alpha cubed we're gonna go from T
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equals -2 to u equals 2 so here the two
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expressions that we want to prove equal
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s from R of T to R a view of PQ and s
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from TD U of P circle are Q circle R so
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let's have a look at the first one so P
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and Q are in here and R of T if you plug
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R minus 2 inside there you're going to
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get minus 3 and when you is to plugged
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in here you get value 5 so going from
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minus 3 to 5 integrating alpha cubed 4
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plus alpha squared so by
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linearity this is really two different
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ones it's alpha cubed together with four
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and also alpha cube together with alpha
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squared so we just use the fundamental
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formula quite simply the first of them
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there's a factor of four coming out and
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then the exponents zero and three so we
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get zero minus three and 2 times zero
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plus three and we have to multiply that
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by five cubed minus minus three cubed
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and then the other term involving alpha
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cubed and alpha squared
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well now the constant term will be 2
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minus 3 over 2 times 5 and we have to
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multiply by five to the fifth now
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because the sum of the powers is five so
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five to the fifth minus minus three to
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the fifth and all of this evaluates to
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minus 3,200 4/5 so with the original
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parameterization of alpha cubed and four
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plus alpha squared we're going from here
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all the way up to actually an endpoint
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of 125 that's five cubed so we're
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talking about the signed area determined
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by these two lines here and the curve
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right now the other side of the change
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of variables signed area from T to U of
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the composition P circle R and Q circle
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R okay R is this expression there it is
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cubed that's the P function and here is
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four plus that quantity squared that's
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inside more complicated but the bounds
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are a little bit simpler okay so we can
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apply the fundamental theorem it's this
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quantity here times the derivative of
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this one minus this quantity times the
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derivative of this one and then well
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it's a bit of algebra to work this thing
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out and then you evaluate it from minus
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2 to 2 and you have to integrate it from
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minus 2 to 2 and there's a 1/2 in front
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and turns out that you also get minus
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3204 over 5 now what makes this a little
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bit interesting
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that if we follow this second
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parameterization it's no longer a by
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action okay so what happens to the
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actual point if we actually substituted
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alpha in here and ran from minus 2 to 2
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what would happen is that it would go up
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like this but because of this this is
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this cubic is kind of doing something
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like this so what happens is that the
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point actually moves along certain ways
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and then doubles back before it proceeds
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to the end point so we are no longer
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talking about a by jek ship no longer
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talking about Reaper amat rising just
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the curve itself we're actually
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traversing it and then looping back and
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so on so it's actually a situation
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that's outside their realm of
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application of a lot of traditional
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change of variables formulas this would
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not be covered but it is covered here
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and it is in fact working out very
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beautifully and now with our last
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example I want to show you a phenomena
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that goes beyond the bounds of the
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previous ones and suggests to us that
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maybe we need to think more flexibly
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about what curve actually is so here the
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example is P equals alpha cubed and Q
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equals alpha that's relatively simple
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just this cubic curve alpha cubed alpha
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and we're going to compose it with 1
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plus T alpha minus alpha squared for the
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bounds T equals minus 1/2 u equals 2
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okay so the one side of the equation is
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s from R of T to R of U of P Q so we're
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going from R of T when T is a minus 1
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plug that in there we're going to get
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minus 2 and when plug in 2 we get 1 so
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we're going from minus 2 to 1 just
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integrating alpha cubed + alpha that's
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signed area well that's just a simple
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application of the fundamental formula
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so the difference is minus 2 divided by
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2 times the sum 2 times 4 and then we're
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going 1 to the fourth minus -2 to the
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fourth for B
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hence the three plus one is four and we
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get fifteen over four
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maybe well circle might okay so that's
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relatively easy we're imagining point
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moving along so when we're at minus two
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we're over here at minus 8 and minus two
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and then as alpha increases up to one
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we're just increasing steadily until we
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get to the point 1 1 so this thing here
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is talking about the projectile going
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from here along the curve nicely to the
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end point there ok now we make this
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change a variable we compose these two
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functions with this quadratic function R
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which is 1 plus T alpha minus alpha
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squared it's graph is like an inverted
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parabola okay so what happens there now
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so now we're just going from T to youth
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so signed area from tdu from minus 1 to
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2 of okay here's R cubed because
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composing with P and here is R to the 1
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okay okay so we calculate this this is
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going to be 1/2 of the Feilhaber
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integral from minus 1 to 2 of this
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quantity times the derivative of the
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second one - the second one times the
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derivative of the first one and when we
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calculate that not surprisingly we get
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15 over 4 as the formula tells us we
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should but what's interesting now is
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that here's this graph of y equals 1
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plus 2x - x squared from minus 1 to 2
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and what you can see is that it doesn't
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just proceed between the values of minus
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2 and 1 it actually exceeds the value it
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goes up to the point 2 when x equals 1
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so when x equals 1 we have 2 minus 1
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plus 1 we're actually at the point 2
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there so what's happening there is that
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if we look at this parameterization here
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okay as the parameter is going from
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minus one to two yes we go from here to
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here but we do so in a rather roundabout
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way when we are at the point alpha
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equals one then inside here we're going
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to get two cubed that means that the
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x-coordinate is actually eight so we're
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going here a lot of the parameters
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ticking along and by the time we're at
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the parameter one the point is actually
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out here and then it comes back to here
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so we're actually traversing a path
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which goes beyond the original curve
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that we are considering and this is part
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of the integration that's being done
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here this thing is integrating their
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curve that goes like this
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out to there and comes back so it's not
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just that we're reaper amat rising that
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actual geometrical segment there we are
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somehow able to access other parts of
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this curve so the idea of a segment
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becomes a little bit more ambiguous here
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but in any case the formula is purely
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algebraic okay so what statement is it
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works no matter which way things are
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oriented no matter what the parameters
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are doing it's just an algebraic formula
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and it's really remarkable all the thing
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so now the question is why is it
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actually true so I think the proof that
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I'm presenting to you now is a novel one
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it's different from the ones that you
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usually find so often arguments in this
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direction involves or geometrical
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thinking in terms of little comparisons
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of small intestinal pieces here and here
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under a transformation
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sometimes Leibniz is you know DX over a
21:56
Y times dy or DZ equals DX over DS that
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kind of argument is used the argument
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here is entirely algebraic okay we're
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only dealing with polynomials and
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everything is really algebraic so it's a
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it's a different kind of argument so we
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want to prove this for
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we'll start with this side signed area
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from tdu of P circle our Q circle R so
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the fundamental theorem tells us that we
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can evaluate this in terms of one-half
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Feilhaber integral from t to you of what
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of this first thing which is P circle R
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times the derivative of the second minus
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the second times the derivative of the
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first fair enough
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and now we're going to rewrite this
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because involved here are derivatives of
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compositions and the chain rule tells us
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how to evaluate those derivatives so
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instead of DQ circle R we now have right
22:54
here D of Q circle R times D R that's
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what the chain rule tells us this is
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equal to and over here the D P circle R
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is replaced with D of P circle R times D
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our ok now next line we're still taking
23:16
1/2 final our integral from T to you and
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now we look at this thing here and
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there's a common factor of D R in both
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terms so I'm gonna pull that outside and
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what do we have left
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the other terms are involving all have a
23:29
composition with respect to R so we can
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write this as P times D Q minus Q times
23:38
DP all of that composed with r so each
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one of these individual things is going
23:47
to have an R so I've inserted where the
23:52
variable is and now we recognize well
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here's p DQ minus q DP that's looking
24:00
like kind of an integral and so we're
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going to introduce a new polynomial
24:09
called n it's going to be the foul
24:11
harbor integral of this thing here s of
24:15
p DQ minus q DP so this is a polynomial
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or a poly number and so we can take its
24:21
Feilhaber integral
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and so what we know then is that if we
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take the derivative of this thing so D
24:31
of n if we integrate and then
24:34
differentiate we get right back to where
24:36
we started so then we just get PDQ minus
24:39
QD P so that means in this expression
24:42
here we can replace what's inside this
24:45
little bracket here with D of n that's
24:51
now D of n circle R so it's 1/2
24:54
Feilhaber integral from TD u of D of n
24:57
circle R times D R but now we recognize
25:03
that this is itself a derivative of and
25:06
the circle R that's just the chain rule
25:09
applied to this so we've reduced it to
25:13
something that's an integral of a
25:15
derivative integral from T to you so
25:18
that's quite convenient all right so
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here's where we are so far we have the
25:23
signed area from TD U of P circle our Q
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circle R equals 1/2 Feilhaber integral
25:29
from TD u of D of n circle R ok but take
25:36
the derivative then an integral like
25:38
this it's just basically giving you the
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original thing evaluated from the top
25:44
number minus the value at the bottom
25:46
number so we get 1/2 of n circle R at u
25:51
minus n circle R at T okay but n circle
25:58
R IU is and of our of you because that's
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what the composition means and here
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we're subtracting n of R of T now we're
26:10
going to rewrite this in this way as 1/2
26:17
of capital s from R of T to R of U of P
26:22
DQ minus q DP because what is n well n
26:26
was the integral of P DQ minus q DP ok
26:32
that's what n is so we're taking this
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thing and evaluating it at R of U and
26:37
what
26:38
subtracting the value this at R of T so
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that's exactly what we mean by the this
26:44
notation of integral from R of T to R a
26:47
view of the same quantity okay but now
26:52
we're in a position where we say hey
26:54
that looks familiar that's just the
26:56
fundamental theorem form for the signed
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area from R of T to R of U of P comma Q
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and so this quantity equals this
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quantity and that's the change of
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variables formula with a purely
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algebraic proof so there's much to learn
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I'm knowing wahlburger thanks for
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listening
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[Music]