okay so you have an integral and inside the integral youll have a function derivative of a function okay and its the chain rule repeated the function no.
dress yeah youre in the middle.
youre usually in the middle its just throwing me off.
|0:32||the chain rule created a function inside of a function.
remember the example i had something like sin(x^3) the derivative of that we got cos(x^3) times 3x^2 right?
so if you had an integral of 3x^2 cos(x^3) dx its not obvious when you look at that
just turns into sin(x^3).
you guys certainly arent at that level of sophistication where you just sort of look and go oh cool thats sin(x^3) +c right?
what you want to do is you want to be able to look at some integral and figure out what the original function was if the integral contains both the function and the derivative.
to do that you use substitution and that helps us..a lot easier.
|1:30||so if for example i had an integral of
thats going to be one of these types of functions cause you have a function and its derivative.
its not exactly obvious what the original function is.
you could probably play around and get there.
but in the beginning it might be a little tricky so what you do we say well whats the simplified function in there?
and its derivative.
well the derivative of x^2 is 2x.
|2:02||i got x so thats close.
so if i let the letter u stand for 5+x^2 can you hear me back there?
can you hear me?
how about now?
so if we let u=5+x^2 du would be well du/dx would be 2x.
so if i cross multiply i get du is 2x dx.
so i look inside the integral and i say well i almost have 2xdx.
i have x dx right? i have x dx so if i if i multiply by 1/2 or divide it by 2 i get 1/2du=x dx.
so now i can move over to the integral
|3:01||and i can substitute so instead of the integral in terms of x
its going to be an integral in terms of u.
but i cant have any x's left.
it has to all be in terms of u.
so i say well this is now the square root of u.
thats easy to do.
easy integral. and x dx is 1/2 du. i put the 1/2 on the outside cause we're integrating not differentiating.
|3:30||if you multiply by a constant the constant just stays there.
so you pull it out of the u make it easier to integrate.
alright this is the same thing as saying a half u^1/2 right?
so now we integrate that.
the integral is 1/2 u^3/2 over 3/2 +c.
|4:01||which, with a little algebra
is the same as 1/3
u^3/2 + c
so now i just substitute back for u
the handwriting gets harder as i move down the board.
thats 1/3 of 5+x^2 to the 3/2 +c.
lets do another one and then you guys get to do a couple.
|5:03||so again i look
now i could take sinx+3
expand it out
its to the 10 so it just wouldnt work
i could distribute cosine
and probably get a whole bunch of messy
integrals so why would i do that?
i look and i say the derivative of cosine is -sin but that wouldnt be very useful cause then id have a- this would be u and this would be du to the tenth which is not helpful.
|5:30||but if i let u be sinx
the derivative of sinx+3 is cosx
and this would now become u^10.
which is easy.
and this would become du so lets do the substitution.
i let u=sinx+3 du is just cosxdx.
so if i have du is cosxdx and u is sinx+3 now i substitute
so usually when you do the substitution with these problems youll end up with a power rule or something simple like a sin or a cos ok?
pretty much all we can give you right now.
maybe a log.
you can integrate a log which is sort of power rule.
okay so this is (u^11)/11
now you subtract -you substitute back
you get sinx+3
to the 11th
how we doing so far? questions?
i dont know what youre doing questions?
im totally confused?
ill just sit here and do instagram? yes?
|7:05||alright so first is you have to figure out
whats going to be u and whats going to be du.
you could sort of do trial and error.
but the idea is you want to find 1 function and its derivative.
so i look here and i say if i let u be cosine the derivative of cosine is -sin thats kind of sin.
but this would have been the u and this would be du^10
|7:30||which is weird.
so i say lets try the other way around what if i let u be sinx well then itd be u+3 which isnt very useful.
but if i let u=sinx+3 du is just cosxdx.
this becomes u^10.
and the remainder is just du.
ive substituted all the parts.
(sinx+3)^10 becomes u^10 and cosxdx becomes du.
|8:00||i dont let u=(sinx+3)^10
thats not helpful.
because the derivative of something to the tenth would be 10 times the thing to the 9th.
i just let it be whats inside the function.
because having stuff to the tenth is a problem for integrating.
right cause you make it to the 11th divided by 11.
its whats inside that i have to fix.
but then i substitute and now i just get u^11/11 and i substitute back.
|8:30||lets have you guys try a couple.
there you go, theres 3 to start off with.
so you look at the first one and you say to yourself what do i do with that 7?
cause thats annoying to have that 7 there.
just take the 7 out.
remember you move constants in and out of the integrand youre supposed to multiply ok?
so let u=x^2+8.
du du is 2xdx.
now i look and i say alright this would be u and i have x left so i just have to play with du for a second.
|9:33||make that 1/2du
so far so good?
so now i substitute and i have 7 times 1/2 times u^5 du.
thats 7/2 or 3.5 if youre feeling decimal.
|10:01||so that becomes
plus a constant.
and then substitute back.
so you would get 7/12.. (x^2+8)^6 +c. to anticipate your question.. no you dont have to turn this into 7/12
|10:30||you could leave it 7 times 1/2 with the thing over 6, thats fine.
simplifying is not important to a question like this.
demonstrate that you know what youre doing.
alright how about this one?
well the 5, yeah you could just take the 5 out.
|11:04||then you could think of that if you wanted
cause lnx/x is the same as lnx -1/x you dont have to do this.
but if it makes it easier for you so then u would be lnx du would be 1/x dx.
|11:34||when you write on the board
step to the side
and see what youre doing
rather than doing this
students find very annoying.
thought id share that teachable moment with you.
in case some of you out of desperation rather than going to med school end up with something terrible like this.
|12:01||du is 1/xdx
so now this is just the integral of
so u is lnx du is 1/xdx so this is just u and the integral of u is u^2/2.
|12:36||and now you just substitute back.
so that would be 5/2 lnx squared.. plus a constant.
which you could write in more than 1 way ok?
so far so good?
|13:00||what about tan6xsc^2xdx?
well if i let u=tanx du is sec^2x dx.
so now this is just
which is u^7/7
plus a constant.
so thats n^7x/7 plus a constant.
|14:05||suppose i had
suppose i had something like that.
cause what am i going to use u substitution on?
|14:30||not going to do u substitution.
first whats 1-cos^2 the same as?
really? you guys know that?
that would just be the integral of sinx.
dont forget your trig substitutions.
because we can make things trickier by first making you do a lot of trig so remember how we did the integral of tangent?
you rewrite that first as sin/cos and its a much easier integral.
|15:04||this is just -cos.
now i could do these as definite integrals.
lets do one of those.
|15:48||i like that one.
|16:05||by the way when in doubt just write pi.
pi..0..1.. 100.. heck maybe one of those will turn up pi is not going to be the right answer to this.
alright thats probably long enough.
|16:31||so lets let u=2-e^x
is -e^x dx
and thats almost what i have.
except for the negative thing.
-du is e^xdx.
so leaving out the limits this would now be negative the integral of du/u.
|17:04||these are called the limits of integration.
now you could change the limits.
ok which we're not going to do today.
change the limits and then you do the whole integral in terms of u and then you dont have to change the ok but one step at a time.
learn how to do this first.
so this is -ln(u)
|17:32||which is -ln of..
now we're going to do that from
0 to ln4.
what is e^ln4?
ok make sure you know that.
e^lnx is x.
|18:00||so this becomes..
you can tell why its plus..minus minus.
so far so good?
okay whats ln1?
|18:31||ln1 is 0.
and the ln-2 but its absolute value so its ln2 which is -ln2.
thats why you have the absolute values.
so i look at this and i say i need to integrate.
we need to do a substitution and i look at the denominator and numerator
|19:00||and i have to make u whats in the denominator.
so i let u=2-e^x.
the derivative of that is -e^x the derivative of 2 is 0.
and then its a minus sign here so -du is e^x so i go over here and now i say okay this becomes u.
this becomes -du.
the integral of du/u is the ln of u not the ln of the absolute value of u.
so then i substitute back and i get the negative log
|19:30||of absolute value 2-e^x.
now i think i have to evaluate that 0 to ln4 so i plug in ln4 and e^ln4 is 4.
and i plug in 0 and e^0 is 1.
so this becomes ln1 and ln1 is 0.
thats why i get that number.
so this becomes the ln2.
thats it my answers the -ln2.
if you really wanna show off, the ln1/2.
why is it the denominator here?
whats the derivative of ln(u)?
so the integral of du/u is ln(u) you should remember some of these integrals.
|20:37||you should know a power.
know the integral of e^x.
|21:01||you should know the integral of dx/x is ln of the absolute value of x.
ill make a handy review sheet of this for you guys tomorrow.
and ill post it somewhere in blackboard.
|21:31||tomorrow i take a 2 hour train trip into the city.
and about an hour and a half trip home from the city.
so i could sit through the entire 3 and a half hours and look at my phone and check out my tumblr but i have a life.
so instead im going to write calculus stuff.
that way thats not off.
so because i dont have a life i will write you guys a review sheet tomorrow, and some practice problems we dont have a practice exam yet.
do that part tomorrow.
ill put up the practice exam
|22:00||probably friday morning.
i will not put up the answers friday morning cause i will be tired and irritable.
cause i wont be thursday night ladies night at the bench no i will be doing calculus.
how do i know about that?
alright so the answers will go up over the weekend.
i dont know if any of you used to go to jake star
|22:30||across the train station. they just closed.
really sad. they say its going to be a new italian restaurant.
so we have a nice big like pizza place across from the train station.
and a new chinese restaurant to order chinese food.
(unrelated to math, inaudible)
|23:02||am i forgetting any? oh sure.
forgot a minus sign.
thats what you get for having awful handwriting.
|23:30||ill erase that and replace it.
so far so good?
lets do some other stuff.
the function f(x)=3x^2+x a find the area of the rectangle.
|24:01||find the area of the curve.
from x=1 to x=9 using 4 rectangles.
right hand rectangles.
b write a riemann sum for the area using n rectangles.
c evaluate the riemann sum.
wanna find the area under f(x)=3x^2+x x=1 to x=9 using
|24:30||4 right hand rectangles.
curve would do something like that.
thats not a good way to draw it.
you want to find this area.
|25:01||im going to make 4 rectangles.
its going to be right hand rectangles.
and at each interval 1 to 3, 3 to 5, 5 to 7, 7 to 9 goes right.
find the height of the rectangle.
its going to overestimate.
too big a number okay the width of each of these rectangles is 2.
the height of the first one f evaluated at 3
you take 3 and put it in.
so 3 times 3^2 is 27.
+3 this is 30.
3 times 5^2 is 75 +5 is 80.
|26:02||3 times 7^2 is 147 +7 is 154.
3 times 9^2 is 243 +9 is 252.
so thats uhh 1,032?
did i get that right?
how about that, right?
|26:31||ok it could be wrong.
now lets write a riemann sum.
riemann sum you have to just sort of think about whats going on but you want to do it in terms of i.
you want to think about how wide each rectangle is.
each rectangle is over 2y each rectangle is now, well i dont know let me take 9-1 divide it by n you get 8/n.
so thats how wide each rectangle could be.
|27:09||the first one is going to be at f(3).
well no its going to be at f(1) +8/n.
the next one is going to be at f(1)+2(8/n).
and 1 of f
the last one
will be f(1)+n(8/n)
well n over n cancel.
that is 1+8=9 which is what you want.
so again how do we come up with this?
i say well i start at the left end point and the width of each interval is 8/n. why is it 8/n?
|28:01||well cause its 8 wide
and im dividing 8 into n rectangles
so i take the first
pint 1 and i move 8/n to the right.
and thats the right end point of the first interval.
and from there i would do another 8/n to the right so thats why its 2 8/n's.
then i move another 8/n to the right and thats 3 and i keep doing this n times.
and the last one you should check the arithmetic should come out for right hand numbers it should come out 9
|28:37||what im writing here is identical to whats over there.
8/n f(1+8/n) + f(1+2(8/n)) + f(1+3(8/n)) dot dot dot
so thats my riemann sum now you just sort of put it in sigma notation its the fun part.
k well thats 8/n so youre always going to have the same thing in your sigma notation.
this will be the width.
that will be (b-a)/n.
|29:31||this will always be i=1 to n.
inside youre just going to have f at the left end point plus the width times i.
k thats what your riemann sum looks like then you take f and replace it with 3x^2+x.
this okay so this is the same as this so 8/n thats the width of each of these.
and i have f and im going to start at 1.
and im going to keep adding 8/n's so im going to have 8/n thats the width each time times i.
|30:33||cause im doing right hand rectangles.
so im going to take the left most end point with 1 but then each interval remember you use the right hand side of it each rectangle is 8/n wide so i have to start at 1 and move 8/n to the right and then another 8/n to the right.
so if were doing left end point rectangles this would be 0 and that would be n-1.
but all our formula
|31:00||the formula works better from 1 to n than from 0 to n-1.
now the last step to get it exactly right is instead of f so i recommend if you do this on the exam if you have a question like this write out all these steps.
youll get partial credit.
so f is 3x^2+x so its going to be 3 x is this. 1 +...
plus x which is
another 1+ 8i/n
so far so good?
now for the really fun part..evaluating this.
|32:03||i know you want to do this.
i dont know if were going to have this on the exam i know if we had to put up a vote i know what the answer would be.
if we do have it on the exam i suggest you do the problem last.
cause its time consuming work on other stuff first do the stuff that youre comfortable with.
these are annoying, i have trouble with them.
okay theyre very tedious.
|32:30||its easy to make mistakes.
but you know when the answer has to come out right?
this has to be the integral from 1 to 9 3x^2+x its x^3+x^2/2 1 to 9 okay so you have to get that number.
so what you do this way you get the same answer as what you get there.
|33:22||so now how do we evaluate this? well ive got 8/n sigma 3|
multiply that out with 1+16i/n
+ 1+ 8i/n
so all i did was foil out this.
wheres the 64 come from?
1 squared 2 times 1 times 8/n 16i/n
which is 64i^2/n^2.
ok lets do some simplifying.
im not going to write the i or the n cause i dont have time for that.
so this 3 times 1 is 3+1 is 4.
3 times 16 is 48 +8 is 65.
|34:30||3 times 64 is 192
starting to get easier. okay?
3 times 1 is 3.
+1 is 4.
3 times 16 is 48 +8 is 56.
3 times 64 is 192 plus nothing...so 192.
skipped a step i dont recommend you skip any steps ok i pity the TAs who have to grade questions like this.
lets make this 3 separate sums.
i= 1 to n of 4 plus 8/n times the sum i=1 to n
|35:31||56i/n +8/n times the sum of i=1 to n of 192 i^2 over n^2|
the sum of i=1 to n
just take the four and multiply it by the n so thats just 8/n
cause thats this formula.
i=1 to n of a constant its the constant times n.
|36:32||the n's cancel
this comes out to 32.
we havent even done the limit yet, oh yeah when you write the riemann sum you have to write the limit as n goes to infinity.
kind of forgot to put that in there.
when were all done we'll take the limit.
ok the middle term we have 8/n sigma i=1 to n.
of 56 i/n
|37:03||so 8 times 56 is 448.
over n^2 so i can take the 8 times 56 and the n times n pull it out times the sum of i=1 to n of i.
which is 448 over n^2 times.. so this one is (n times n + 1)2
|37:35||that simplifies to 224
one of the n's cancel
and im left with (n+1)/n
we're not done.
now ive got
times the sum of i=1 to n
8 times 192
i could just be making this stuff up and you could just write it down.
lets see thats 1600 minus...yeah 1536.
ok and ive got the sigma there, sorry
|39:06||try to be less sloppy.
and that is 1536 over n^3 times the last formula n(n+1)(2n+1)/6.
|39:32||which simplifies to 256 times (n+1)...
alright take the limits.
the limit as n goes to infinity of 32 is 32.
well the limit as n goes to infinity of n+(1/n) is 1.
and this ones going to be 24.
|40:00||and this one
comes out to 2.
k so therefore thats the sum is 32 +224 + 512.
which is 768.
|40:32||now i wrote a whole bunch of this stuff out
and put it on the website last week on blackboard.
did any of you guys look at it yet?
i recommend you look through it cause i basically work through one of these step by step.
i write everything out nice and easy well 1536/6 = 256 the limit as n goes to infinity of this is 2.
|41:00||ive got a 2 over 1 okay?
so thats why i had 2 at the end.
how do we feel about this stuff?