|Start||so derivatives and integrals
so this is the first fundamental theorem of calculus.
so if you have an integral from... 3 to x i should do 0 to x because they like 0.
f(t)dt okay and thats g of...g of x.
then g'(x) would just equal f(x).
|0:32||right? we did this the last time.
and g double prime of x will equal f'(x).
so say i told you that we had something like this
|1:08||and i wanted to find g'' at uhh...
suppose i wanted to do something like that.
would be (1+cos^2x)
times the derivative of cosine
which is -sine.
so if i wanted to find g'(pi/3) i now plug in pi/3 right?
would be 1+cos^2 of pi/3
so cosine of pi/3 is a half so cosine squared would be a quarter.
so this would be 1+1/4 times minus radical 3 over 2 and im not interested in what that is.
and then if i wanted to find g''(pi/3) i take the derivative of this which is product rule.
|2:33||so i have g'(x)
times the derivative of -sin which is -cosine.
plus -sin(x) times the derivative of 1+cos^2 which would be 2cosx to the 1 times -sin(x) this is very tedious.
|3:00||k? and thats..
this comes from the product rule.
you look scared.
alright and then you just plug in again.
so g''(pi/3) would be this.
|3:33||plus minus radical 3/2
2 times a half
and you simplify it.
so thats a typical type of thing you could ask.
cause were testing if you can see the connection between the function, the integral, and the derivative.
the derivative of the integral okay?
|4:00||so when you do the derivative of the integral the integral kind of goes away.
especially if you just go from 0 to x or any constant to x.
so far so good?
so we could give you..i'll wait a minute.
i'll wait until everybodys got that down.
you know it'll be on the video later.
alright so another thing you could do with this fundamental theorem stuff
|4:30||i did some of last time
and i tell you this is f(x)..f(t).
|5:06||i give you something like that so we did this last time.
so remember what this means.
this means g(x) basically means the area until you get to x.
so g(4) would be the area up to 4.
g(6) would be the area of 6 which would be this area minus that area.
right g(8) would be the area from here to here.
and so on.
so if i say where..
|5:30||is g increasing?
where is g increasing?
|6:00||g is increasing..well
remember from last semester g is increasing when g'
and what is g'(x)?
g'(x).. is f(x) so wherever f(x) is positive g' is increasing.
so that would be.. (0,4) and (6,8).
|6:31||that make sense?
i want to know where is g increasing?
and to know where g is increasing i need to know where g' is positive.
well g' is just f.
so f is positive between 0 and 4 and between 6 and 8.
so what if i said where does g concave up?
or concave down it doesnt really matter.
|7:11||so g is concave up where g'' is greater than 0.
so g'' is greater than 0 well if g' is f then g'' is f'.
|7:32||and f' is just the slope.
so that would be from here to here and from here to here.
thats the 2 places where f has got a positive slope.
so 0 to 2 and 5 to 7 that make sense?
i like these kinds of questions i didnt put them on the exam last semester but
|8:02||they're kind of okay.
some people really like them some people dont like them at all.
the AP loves them.
theres one on every AP exam.
okay we feel we understand that stuff?
we're going to learn something new now i just want to make sure everyone feels okay about this.
yeah? thinking about chemistry?
|8:30||no chemistry. no chem thoughts.
we're going to learn something new.
substitution rule, same thing.
its really a method more than it is a rule.
what are the 4-what are my 4 favorite rules?
product rule chain rule quotient rule
but not in that order.
so substitution so now we're going to start to learn how to do hard integrals.
so we've really only been doing very simple integrals so far.
|9:31||you need the substitution rule..substitution method
when you want to work backwards from the chain rule
we havent seen
this is what you use when you havent a clue essentially theres 2 things in it
and one of the things in the integrand
is the derivative of the other thing in the integrand
so for example
say you have the inegral of
i think the webassign tries to teach you how to do this.
|10:02||but i'll do my flavor of it.
my version of it.
so how would you do the integral of this?
well one thing you could do is you could expand this out.
to the tenth.
and then distribute the x^2.
so that would be easy if this was say squared because then you could just sort of foil it out.
but the 10th is a bit messy.
but then you could notice something.
x^3.. the derivative of x^3 is kind of like x^2.
because you just, its 3x^2 right?
|10:32||so if we let u
we like to use the letter u.
equal 5+x^3 and du dx the derivative of u would be 3x^2.
there will be a reason for this in a second.
and cross multiply this you get du.. is 3x^2dx
|11:01||so 1/3du is x^2dx.
so far so good?
why take out the 1/3? alright i look at this integral and i say im going to substitute things.
so i have 5+x^3 so if i let this be u this would just be the integral of u^10 which is trivial..okay?
but im stuck with an x^2 and a dx
|11:30||so i take the derivative and i say
ive got 3x^2 dx's.
but i only want to substitute for x^2dx so if i move the 3 to the other side now i can go over this integral and i can say this is u^10.
and this.. is 1/3du.
pull the 1/3 out and thats an easy integral.
|12:05||so far so good?
so whats the integral of u^10? u^11/11.
so thats 33 down there right? because youve got a third and you got 11.
you all see down there?
and then substitute back so you would get 1/33(5+x^3)^11+c
|12:35||okay? and the 33
comes because this is 1/3 times 1/11.
and double check that in your head so take the derivative you get 11 thirty thirds (5+x^3)^10 times 3x^2 so notice right? thats a third that cancels with the 3 and you get the x^2 and the 5+x^3.
so what the u substitution does
|13:01||is it undoes the chain rule.
it basically looks at this integrand and says theres a chain rule going on in here.
so i'm gonna work the chain rule backward.
so lets do another example.
suppose i had sin(x)
|13:54||alright so i would do u substitution.
so im looking for something and its derivative.
|14:00||now i look at-so the derivative of sin is cosine
and the derivative of cosine is sin
so how would i know which to do?
well this is under the radical so thats causing me more irritation.
so i'm going to let that one be u.
and let u equal 1+cos(x).
and the derivative would be -sin(x) which is almost the same thing as sin(x) right?
so if i cross multiply
|14:30||i would get du
and tonight i'll put the pages up from my book that do the u substitution.
you can look at lots of examples.
so now i come over here and i say well alright i could make this square root of u and this would be negative du.
because du is -sin(x)dx so negative du is sin(x)dx.
who is confused?
alright so the idea is you want to look at the integral and you want to put u into one part and the rest of the part has to be some form of du.
which usually you play with the constants.
so you can turn this now into a very simple integral.
so i look at the beginning and i say ive got the square root of 1+cos(x).
and i got sin(x).
i know the derivative of 1+cos(x)
so this is basically -sin(x) just off by a minus sign.
so if i let u be this and then du is -sin(x) then i can substitute everything in here.
and make this remember this is an easy integral right? this is just u^1/2.
we'll do like 3 more of these and then youll get good at it.
that becomes u^3/2 over 3/2 plus the constant and then you substitute back.
see whats going to happen in integration is youre going to look at integrals and youre going to say i have no idea what youre doing and you start trying out all of your tricks.
and see if you can come up with one that will work.
what happened to the minus sign?
so this is-my du is -sin(x).
but ive only got positive sin(x) in here.
|16:32||so when i substitute i use negative du.
you get rid of the minus because this is minus here.
right? so -du would be positive sin(x).
okay? thats why i have the minus sign here.
lets do some more of these after about 3 or 4 you'll start to get the hang of them.
or maybe not.
so the integral is xcos(x^2)dx.
|17:13||alright so lets do some substituting.
well i see the cosine i cant really i cant do much with the cosine.
if i said i'd have u equal cosine the derivative is sin and there isnt any sin in here.
but if i let u=x^2 the problem is i could do the integral of cosx.
i could do the integral of x or some x^2 but the cosine is sort of messing things up.
so im going to let u=x^2 and then du dx is going to be 2x.
why, why do i want to do that? well i want to get rid of this problem.
now i look at the integral and i integrand and i say i've got x dx so if i cross multiply
|18:01||i've got 2x
i only want to substitute for x dx so i just divide by 2.
k now i go here and i substitute.
i substitute for cos(x^2) and i get cos(u)
|18:30||and x dx
is going to be 1/2du.
put the half outside i could put the 1/2 inside but i generally put constants outside of the integral its easier that way.
alright whats the integral of cosine?
soo.. u is x^2 so this is 1/2 sin(x^2) plus c.
and if you took the derivative of sin(x^2) what would happen?
|19:01||you'd have cosine of x^2
so thats where that x appears from.
so u substitution is working this backwards over and over again.
so lets give you guys one or two to practice.
nice easy ones.
i know its new for almost all of you so i wont give you killer ones yet.
|19:31||thats for the second half of the lecture.
believe me you guys will get the hang of this very soon its really not that bad.
theres your first 2.
give you guys a couple minutes.
so i look at this i see sin^4xcosx.
|20:30||so im going to need u substitution.
and i say to myself well if i let u=cos the problem is du would be sin sort of and that would be du^4 which doesnt make any sense.
right? im only used to one dx at a time or one du.
so du^1/4 would be sort of painful.
and if i let u=sin^4 the derivative of that is 4sin^3cos thats not what this is.
but if i let u just equal sin
|21:02||then du would be cosine.
instead of writing the dx down here im just gonna cross multiply right away.
you can skip that step if you feel the need to.
and now its easy to substitute i say well this is u^4.
this is du.
|21:30||so thats now u^5/5.
and i substitute back and i get sin^5x/5 plus c.
how did we do on that one?
of course its easy if i tell you to use sin.
so in the beginning the mistake people make is theyre not sure which one or theyre trying to substitute for the whole expression which you dont need to do you dont need the power.
because you can always do u to a power
|22:00||so dont worry about substituting for the power.
but you understand the idea?
so some general rules if you have something to a power and you have something else thats not to any power just the power of 1 still a power okay?
then the thing with a higher power is almost always going to be u.
but usually you can ignore what the power is.
the second thing is you have 2 things in the integral and theyre 1 power apart
|22:30||x^10 and x^9
then the higher power will be u and the lower power will be du.
well do an example of that in a minute.
alright what about here?
i see cos(e^x) and i see e^x.
so if i let u=cos theres no sin in here.
but if i let u=e^x du is e^xdx.
|23:00||so this now just becomes integral of
thats cosine of u
you like that one?
so thats just sin of u plus a constant.
so thats sin... e^x plus c.
by the way some professors are very fussy about the whole plus c thing.
they take off points if you dont write the plus c.
|23:32||so try to remember it.
when youre done, when youre done with your exam look at all your integrals and dont forget to write, just write plus c after all of them.
but not for the ones where you take the derivative of the integral.
okay? dont mess that up.
lets do some more of these.
|24:10||so remember what i told you, you look for something wrong
and if theyre 1 power apart
the thing thats to the higher power will be u
and the thing thats to the lower power will be du.
okay? higher power will be u lower power will be du.
|24:30||lets let u=6+x^10.
why do i choose that?
because the derivative of x^10 is something x^9.
and the derivative of 6 is just going to go away.
so du would be 10x^9dx.
you all see why i can put the dx over here right?
its du dx and then i just multiply across.
so now i look at the front
|25:01||and i say well this
wont become u^4
so i just have to play with this a bit.
so i have 10x^9 so i have to divide by 10 and get x^9.
and now i substitute.
this is u^4 x^9dx is 1/10du.
|25:32||so thats 1/10...
plus the constant.
so u is 6+x^10 so this is now (6+x)^10 over 50 plus c and yes you could have written 1/10 and the 1/5 separated thats fine.
howd we do on this one?
|26:01||starting to get the hang of it?
good should i kill you with some hard ones now? sure.
oh yeah im sorry i wrote that wrong.
thats what i meant.
what should we let u equal?
also known as log.
why is it so hard to look at the letters of ln and say log?
i mean lots of things in english arent not pronounced the way they are spelled.
none of you can pronounce february correctly so.
some of you can pronounce february correctly.
most of you say febUary and thats not the way its spelled so there you go.
so the derivative of u
|27:30||is 1/x(dx) and look
i have 1/x(dx) thats this.
right? that is 1/x(dx).
so this is just sin of u.
integral of that is -cos(u) plus a constant.
and then substitute back -cos(lnx)
|28:01||plus a constant.
how did we do on that one?
yes no lovin life?
not lovin life?
time to give you one slightly harder.
now try u substitution.
lets let u=cosx now
|29:01||how do i know to let u equal the denominator?
well if i let u=sinx i'll have du in the denominator and that doesnt do me any good.
i dont know what to do with the du down there.
its got to be up in the numerator.
but if i let u=cosx du is -sinxdx.
so i can now rewrite this as -du/u.
|29:31||and that is ln-yes?
thats a very good rule.
so yes if you have a function over a function u will be in the denominator it cant be in the numerator.
because if u is in the numerator du above almost always.
okay? because here right?
|30:00||sort of was in the denominator but not really.
theyre really multipled this is really sin of lnx times 1/xdx.
but generally you cant have du in the denominator because theres nothing you can do with that its nonsense.
so you have to have it up in the numerator.
alright so this is.. -ln(u) plus a constant substitute back.
|30:31||i suggest you memorize
what the integral of tangent is.
so one to add to your pile and you should be able to figure out cotx once you know tanx.
how did i get from here to here?
well whats the derivative of ln?
1/x so this is 1/u.
right? so the integral of -du/u right? is the same as -1/udu
|31:32||so thats why its natural log.
heres one thats annoying.
so we'll do this one as a team.
cause this one is slightly annoying.
what if i let u=tanx?
well du would be sec^2 which doesnt do me much good because ive got sec^10.
|32:01||what if i let u=sec?
whats the derivative of secant?
sectan, so what if i rewrote this?
as sec^9x(secxtanx)dx i just pulled out one of the secants.
we can be quite nasty if we need to be.
we've got even more evil ones up our sleeves.
du is secxtanxdx.
so far so good?
once i showed you.
well you'll get there on your own.
so this now becomes u^9du.
|33:01||which is u^10/10.
and thats.. sec^10x/10+c.
i have one more type to show you guys.
|33:41||suppose you have something that looks like that.
now if i let u=x du is just dx.
that doesnt do me much good.
what if i put the x back under the radical?
|34:01||then i'd have x^2 and x^3 so that doesnt do me any good.
if i let u=1+x what if i let u=1+x?
then du is just dx right?
what do i do with this x?
well if u is 1+x then let u-1=x.
and now this would become (u-1) times the square root of u du
|34:44||i look at this i say if i let u=x
du is dx
which is just 1 that doesnt do me any good.
but if i let u=1+x so this will now become the square root of u.
this is just u-1 now i could just distribute that square root of u.
|35:01||this will become u^3/2
so how do i know to do that?
theyre the same power.
i look and i say if you basically have an x and an a+/- x thats the trick that you use.
if you look at the integral and you have one of them is some constant +/- x or the other way around x+/- a constant and the other is just plain x
|35:30||this is the trick that youre going to use.
the whole next month is going to consist of techniques of integration.
its like you have little tools in your tool box figure out which one to use.
so right now you've got a second tool.
we're going to do integration by parts parts by fractions we're going to have about 5 tools when we're done.
then the trick is just which one do you use.
my advice is you try one after the other until you get one that works.
okay so this you can integrate see here if you distributed the x at the beginning
|36:01||you would have had this
and that you cant actually do anything with.
but here you can.
cause this this is just u^5/2 over 5/2 minus.. u^3/2 over 3/2 plus c and then substitute back.
so you get 2/5 (1+x)^5/2-2/3 (1+x)^3/2+c.
|36:50||teachers like that one.
sort of an advanced u substitution one.
i know i had it when i did integration back in the day.
|37:01||taught it right here at stony brook.
you guys have a chem exam tonight?
alright so we'll stop a few minutes early.