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How to find the integral of base other than e functions
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Feb 19, 2025
In today’s lecture , the students and I were working on finding integrals of base other than e functions. We worked on several problems with different wrinkles
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basis other than e
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um applications right so here what we
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have is this I gave you the general form
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right a to the U is 1 L of a a the U
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plus C now here this is pretty simple we
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just have to apply the formula straight
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up but now there's a difference here I
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have
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4X DX so I have to use what my use
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substitution first I can't just do it
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like this I have to use a use
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substitution okay because if you were to
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just so much you may be tempted to say
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this is equal to
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one this is equal to 1 4
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right * 4x + C that's not true we can't
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do that because it's a negative so it's
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not right so what we do is we have to
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use the U substitution so we can get to
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this point so I'm going to let you
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equals what wait M I have a question yes
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sir for uh the first part of D1 where it
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says 3x DX we do the thing where it's
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like Ln 3 * 3 * 1 or would we do 1 over
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ln3 right here yeah you can stop here
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and that's it you're done you don't
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would we do that or would we do the one
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ln3 * no I'm finding the derivative here
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so it's asking find the integral yeah
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here we find the integral okay so that's
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not the derivative this is the
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derivative okay and this is the integral
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first you want to find the derivative to
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find the derivative is Ln of a over a to
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the U * D over DX or if you don't like
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this uh way it looks here you can show
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Ln of a * a u * U Prime and then we did
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the example 3 x Prime is Ln of 3 * 3 x *
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da of X which is 1 and we use this the
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same thing here 3 X2 - x right the
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derivative is Ln of 3 which is Ln of a *
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a to U which is 3 X2 - x * the
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derivative of that which is 2x - 1 so we
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don't need use that to find an Ral no
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okay right now for the integral this is
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the new formula a u is 1 n a * a u + C
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right so now we have a problem we have
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4X we can't just apply this formula
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straight up we have to use a
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transformation we have to use what the U
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substitution right so I'm going to let U
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equal
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whatx and du
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is DX negative DX or 1 DX right and then
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so now we can go back and see this is -4
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to the U duu and now we can apply
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that right so now what is
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it L4 time don't forget your negative
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right
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l u which is 4 to the
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what X plus c c all right so you see
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what we did here so you have to go an
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extra step you have to when this is not
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just like straight up four to the U
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eight to the U you have to use a u sub
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actually it's it's the same result but
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you can't just do that like that yeah
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it's not the same because before that we
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didn't have a negative remember oh the
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negative was missing that's so it's not
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technically the same result right so now
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let's go on your sheet and we're going
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to let's go on U your paper and I want
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you all to work on U there's a problem
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there
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x² so how we going to solve this one
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here we have
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X2 + 2x DX we trying to find the an der
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so what are we going to do
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here you can split it out split it out
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good way to start so we have x² DX right
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plus
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2X DX so what's the antiderivative of
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that uh
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X over three right now here because
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you've already done it you already know
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what that's going to be
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[Music]
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right2 2 c 2 X+ C because the outcome is
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going to be the same as the last one if
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you use a u substitution you going to
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get to the same thing just remember that
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right now let me challenge you all a
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little bit what if you have
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this two to the
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x² 2X and then X
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DX I want you all to try and work with
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this I give you about 5 minutes to come
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up with the solution so we have 2 x² X
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DX I want you to find the an of
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that
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for
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got it yeah all right
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so what did you do I just let = x² let =
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x² we all good on that du will be what
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du is 2x DX all right 2x DX and then and
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then you oh and 12 du is X DX X DX right
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and now go back here and just replace it
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so this is two to
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what to U du du and now we can find of
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that right yeah so that's right 12 right
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time
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2 2 U
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is M plus plus C all right so you see
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you have to use your Transformations you
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have to be very careful to recognize
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that there's a relationship between this
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and that okay
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I'm I'm not reason it I'm keeping it
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there so here we use this U substitution
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again we went let U = x² okay du is 2x
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DX Y is D oh it and then X DX is 12 duu
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right so if you go back and you can
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substitute x² by U and then your 1/2
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cannot you canot forget that so this is
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now
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and then you know that thetive of 2 the
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U is 1 l a a u plus C so we justl it in
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there right now let's try to see if you
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can work on this one here this is more
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challenging 3 2x 1 + 3 2x DX so I want
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you all to go ahead and try to work on
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that on your own like
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twice um you have several options you
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can use like different different things
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here so whatever you you think going
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work for you all
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right so you have two options so let me
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know what you want to
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do all right so
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basically
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hold
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dig
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dig oh he likes to dig all right so you
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see so we have 3 2x right over 1 + 3 2x
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DX so we know automatically that there
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going to this is going to involve a you
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substitution right Kevin yes so now what
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I want to do is this I want to see if I
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can establish something between this and
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this here right so I know by definition
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that Dera of 3 2x is natural of 3 * 3 2x
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*
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2 right which is 2 natur of 3 * 3
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2x now I know that if I let u = 1 + 3 2x
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the derivative of U will be 2 natural of
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3 * 3 2x DX
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right the derivative of
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that so what's the derivative of 3
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2x uh we found it out there yeah right
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there right so if you find the of one is
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zero right so this is 2 natural of 3 * 3
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2x DX which is what we trying to get get
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to we try to find something that so that
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we can establish a relationship between
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this and
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that right so now I know that 3 through
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2x DX is Du over 2 natural I'm just
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trying to solve for this alone so I can
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substitute that up here okay you see it
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now so this is where I got the one/ two
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natural three from you you put at the
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bottom bottom cuz I'm I'm trying to find
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this by itself right yeah so now now
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that I find this by itself I can take I
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can put this one
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outside right because this is one two
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natural 3 * du I can put it aside just
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like we've been doing the entire time
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the only reason why this looks a little
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weird is because we have a natural
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LA right and then now that gives me what
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du over U oh you see it and now I know
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what's anti- one over U is natural of U
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plus C and I don't forget to add this to
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it I'll do
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you and then I replace this by 1 + 3
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2x do that make sense now yeah all let
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me get so the separated Ln of absolute
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value of U that's like what we did like
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before before yeah and then you replace
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You by original value so I'll give you
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another one that's similar to this and I
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want you all to do it so let's say we
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have wait hold on let me uh yeah go
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ahead and write that so what if we
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had
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all right so let's do this you guys go
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ahead and do this one
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2X over 5 + 2x DX I want you to find the
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an using the same method that we use
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here
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