|Start||from 0 to pi/2
whats the sin of pi/2?
and the sin of 0 is 0.
which equals 1 get the idea?
these arent so bad this is just like when you do derivatives you say great why did i do all that weird stuff?
i say well, now you understand it.
so i showed you how to do a bunch of these indefinite integrals before.
but lets practice a couple.
so this is called a definite integral so an indefinite integral is when you do an antiderivative with a plus c.
and a definite integral is when you plug in the numbers.
|1:11||write a little dx there
when you do integrals.
dx has a purpose for now just sort of remember that its there.
kind of like when you took a derivative you have dy dx sort of appears when you do the integral you have the dx which is
|1:31||stays on the derivative side
and when you integrated it disappears
so this is called the indefinite integral.
so this would just be x^3/3 minus 3x^2/2 plus c, notice the dx is now gone.
in the process of integration you get rid of that youre going to need that dx for stuff.
which we'll see... maybe next week, certainly the following.
|2:01||and if i wanted a definite integral lets say i was going from 1 to 2
well then why is it called indefinite well cause
theres an infinite number of possibilites.
just depends what you pick for c so far so good?
so the definite integral is the same antiderivative
|2:32||but now im going to evaluate it from 1 to 2.
so im going to plug in 2, im going to plug in 1 and im going to subtract.
so this would be 2^3/3 -3 times 2^2/2 minus..uh oh.
|3:00||minus 3 times 1^2
and you can figure out what that is.
i dont really care.
its messy. ok but you can figure out what that is.
ill leave the fractions for you guys youre all going to be doctors youll be excellent at this stuff.
|3:33||lets practice a couple make sure everybody is good at this.
|4:21||okay lets see if you can do that one.
|4:36||okay so we're going to find the integral
from 0 to 3
you always say from the bottom to the top.
of x^3-6x dx so this is x^4 over 4.
-6x^2/2 which is 3x^2.
fron 0 to 3
|5:02||if you plug that in you get
minus..that parts just going to come out 0.
okay and if you do this you should get minus uhh 27/4 or -6 and 3/4 so how can that be negative?
well if this is giving you the area under the curve what happened is
the curves doing something like that.
thats the square root of 6 and 3 is somewhere here so this area underneath is bigger than that area above.
so this is another thing i should tell you about this is whats called sign
so the area above the x axis but under the curve comes out positive.
and the area below the x axis so kind of above the curve comes out negative the negative part turned out to be bigger than the positive part on this curve.
okay, on this area.
so for example suppose we look at our favorite function sinx
|6:44||what do you think we would get if we did the area from 0 to 2pi?
we should get 0.
cause the part above the x axis should cancel with the part below the x axis.
so lets do that integral.
the integral from 0 to 2pi
|7:00||of sinx dx
from 0 to 2pi
okay cos2pi is the same thing as the cos0.
you get -1 minus -1
|7:32||which comes out 0.
so thats what wed expect.
so by the way sometimes on these kind of calculus problems if you know what the graph looks like you can just get right to the answer you dont even have to do the evaluation.
so sometimes by symmetry or by knowing geometry of whats going on youll be able to get to an answer very quickly.
lets do another example of that
|8:01||so you see what i mean.
so we'll do the area under y=x, f(x)=x
|8:30||from 0 to 6.
but what would that look like?
thats just a triangle right?
y=x is a straight line goes from the origin itll go to point (6,).
the area would be 1/2 times base times height you get 18.
now lets make sure that our calculus agrees.
|9:10||so if i did the integral from 0 to 6 of x dx
okay thats x^2/2
from 0 to 6
which is 6^2/2
minus 0..is 18 thats exactly what we thought we were going to get.
okay so sometimes you should be able to look at this and figure out exactly what the area is
|9:32||other times it wont be so easy.
lets just do a couple more make sure we understand everything
|10:32||so if we do the integral from 1 to 9 of
+ t^2 radical 2
all over t^2 now of course t...x..the letter doesnt matter right?
lets see if you can do that one.
|11:24||its the integral from 1 to 9 the top if 2t^2 plus..|
|11:32||t^2 times the square root of t
all over t^2.
soo how do we do this? well first of all youre going to want to break this up.
integrand as its called you do the integral from 1 to 9 of 2t^2/t^2 + t^2 radical t / t^2 -1/t^2
which then simplifies to integral from 1 to 9 of
2+ the square root of t which is t^1/2
-1/t^2 which is t^-2
to get to this stage-yes?
you wanted to bring the t^2 up?
and make it t^-2?
|12:30||you distribute it youre going to end up at the same spot.
so its unnecessary work but sure.
but you know its sort of doing it the hard way.
but that doesnt stop lots of you guys.
you generally want to do this as efficiently as possible but you dont have to.
when you see multiple things on top and 1 thing on the bottom usually we do that to confuse you and see if youre clever enough to break them into the individual things over the denominator.
|13:00||so we do it the other way on purpose?
alright whats the integral of 2?
its going to be 2t.
the integral of t^1/2 is going to be t^3/2 divided by 3/2 and the integral of t^-2 is going to be t^-1 over -1 all evaluated from 1 to 9
|13:31||or if we want to simplify that
itd be 2t
1 to 9
so far so good?
you know i got my microphone back i got an email monday night from the audio visual people i guess somebody actually comes around at the end of the day
|14:03||closes things up
and say hmm the uh microphone
disappeared from your classroom
one of you people is bad
so i emailed back and said well at 4 o clock it wasnt here.
so i got the very nice email saying oh thank you now we can narrow down the suspects so i guess they found the person.
they probably killed him or her the head is probably mounted outside the SS its a lesson for those of you
|14:31||just a certain example for others
alright we plug in 2
we get 18
okay square root of 9 is 3, 3 cubed is 27
2/3 of 27 is 18
so thats 36 and 1/9
|15:01||36 and 1/9 would be
and thats 3 and 2/3
which we could turn to ninths
and that would be 33 ninths
and that gives you 292/9
am i close?
|15:31||i dont know
howd i do?
do i get to go to college?
thats why i stand in the front alright soo i'll screw those up though and give you a chance to laugh at me i promise.
this is the third time around.
you should know by now right?
yeah you youve seen me do this before.
lets give you one more of these.
lets give you a really entertaining one.
these are really fun arent they?
much less painful than the other stuff.
im going to make it painful now
|16:40||okay who needs more time?
got it? whats the answer?
|17:11||who needs more time?
whats the integral of e^x?
thats the easy part.
|17:32||no plus c.
okay so this is e^x so we can get you with the logs so you gotta be careful.
thats e^ln5-e^ln3 whats e^ln5?
there ya go.
|18:01||that wasnt hard.
you should be able to do that one in your heads, as i said, alright.
e^ln5 is 5.
e^ln3 is 3.
okay make sure you know that e^lnx is x.
ln is pronounced log.
okay lets give you a more serious one.
im looking for ones in the book.
|18:34||oo this is painful.
then i'll go back and do some other stuff.
i think ill give you guys about 3 of them for practice right now.
|19:02||good i want you guys to be good at these.
heres 3 to work on.
alright so the first one says this one says the integral from 0 to 2 of y-1 times 2y+1 dy.
this one says the integral from 0 to pi/4.
secant theta tan theta d theta if you can read that one.
|19:33||this says the integral from 1 to 2
all over x^4
you all read those?
lets do these.
so how do we do the first one?
well what i would do first is i would foil this out
|20:00||so this is
you multiply that out correctly?
that becomes 2y^3/3 -y^2/2 -y.
0 to 2 16/3-2-2
it comes out
as they say in the south shore.
alright that was easy.
what is the integral of secant theta tan theta?
secant theta because the derivative of secant theta is secant theta tan theta.
its going backwards.
is secant theta
from 0 to pi/4
now comes the fun part whats the secant of pi/4?
if you dont know at least on your exam write secant of pi/4.
get yourself some partial credit.
might even get full credit depending on the professor.
the cosine of pi/4 radical 2/2 you turn it upside down you just get the square root of 2.
now that weve got the square root of 2 -1
if you do it on your calculator its about .414.
ok why is the secant of pi/4 radical 2? well you take radical 2/2 and you turn it upside down you get this equal to that if youre not sure cross multiply okay.
we know what the cosine of pi/4 was its 1 over the square root of 2.
good what do you get when you turn it upside down?
square root of 2.
well its also that.
its 1 over the square root of 2 or square root of 2/2 depending on who your teacher was in high school.
use that triangle, okay.
|22:31||as long as you get the right answer thats what matters.
some of you memorized the unit circle.
some of you still sort of do it in your head.
i could put that back up on my insta that was on the insta already yeah its in there.
that stuffs on there from last semester i could delete it and put it back so you could re like it.
im the old guy who puts up math equations and gets 100 likes.
i dont know what that says about my life.
but what gets the most likes?
hello the puppy pictures
shes a little older than that now so kinda sad.
okay so we could break this up and we could make this its the integral from 1 to 2 of x^3/x^4 which is 1/x + 3x^6/x^4 which is 3x^2 dx
ln of the absolute value of x
from 1 to 2 we dont really need the absolute value because the absolute value of 2 is 2.
but you can write it if you want.
and you get ln2 +8 minus ln1+1 whats the natural log of 1?
|24:02||0 because the ln1 is always 0.
so you get ln2 +7 or about 7.7.
got the idea?
so one of the fun things about these antiderivatives is they stand for the area under a curve.
but integrals also have other meanings an integral generally is just a sum
|24:31||so how many of you have taken physics?
good so you should start using integrals in physics it can get really hairy when you get to physics 132 132?
126 youll see it in there
|25:01||but ive been here longer
he came here in 1965
but i came here in 1961
course i was
a baby but
professor lee has been here a long time
he used to be the tennis champion at my tennis club i used to play him when i was your age and he was my age.
the age i am now.
how is he doing?
hes fine okay good answer.
|25:31||dont want to get yourself in trouble youre worried ill call him up and tell him.
i would never do that.
well this is a secret no one even knows what were saying right now.
no its not like were being recorded or anything.
but if youre doing physics you learn that theres displacement and theres distance.
so displacement is the changing of position where were you when you started and where were you when you finished?
so in physics theyre allowed to ask questions like so you get up in the morning
|26:01||and drive to school
and stop and get a bagel
you go out to lunch
you come back to school for a little while
go to your job
and come back home.
whats your net displacement? your net displacement is 0 because youre back where you started.
but the distance is that whole route ok?
so integrals can be used to find either displacement or distance.
so displacement is easy for distance you have to factor in
|26:32||that you might be going different directions at different times.
so suppose we have an object that moves along a line and were given the velocity
|27:12||find the displacement
that says 15 if you cant read it
find the displacement in the first
in other words
|27:30||whats the total displacement from time t=0
to time t=6?
so all you would do would be to integrate from 0 to 6 t^2 -8t +15 if you integrate velocity you get position ok?
because if you take the derivative of position you get velocity so if you integrate that the antiderivative you can figure out how far this thing has gone.
so that comes out t^3/3-8t^2/2 +15t.
from 0 to 6 that is equal to
from 0 to 6.
that is equal to 6^3/3 -4(6^2) +15(6) thats the first one the second one when you plug in 0 youre just going to get 0.
you can ignore it.
so lets see thats uh
|29:02||162-144 thats 18.
so thats the net displacement of this object from time t=0 to t=6.
so heres the fun thing since lots of you havent taken physics velocity is a vector what that means to put it in simple terms is
|29:30||if youre going 1 direction its positive if you go the opposite direction its negative.
okay soo you can say for example to the right is positive to the left is negative.
up is positive down is negative it doesnt really matter.
ok but theres a difference between going this way and going this way so this object at velocity t^2-8t+15 if you factored that
|30:08||and you put that on a sign
remember this from last semester?
and you found the value of velocity and youll find it goes like this and the object its kinda doing something like that.
|30:30||the tip you get t=3
its going to the right.
then it turns around and goes to the left so t=5 and its going to be somewhere around there t=6.
okay so notice part of the time its going to the right part of the time its going to the left and then we found the displacement is this.
its where you started versus where you finish.
okay thats this is actually on the x axis so the displacement it says 18
|31:02||so youre 18 units away from where you started.
but the total distance traveled is not going to be the same.
and the distance you break up the integral.
because when its moving to the left you get negative values and you dont want to subtract those off when youre doing distance.
so if we were doing the distance instead of the displacement you need 3 intervals you would want to go from 0 to 3 of t^2-8t+15.
|31:37||and then because were going to the left you want to subtract the answer we would get from 3 to 5 and then you would add from 5 to 6|
|32:01||does that make sense?
this ones a little more tedious obviously okay? yes?
im subtracting the second one because velocity is negative i would get a negative answer and i want to make it positive.
that make sense?
so the simple rule you say to yourself is wherever velocity is negative i just subtract the integral i know most of you are not taking physics but
|32:32||you get to do this in 126 anyway
so you want to break up
find out when its positive find out when its negative so lets just do one.
something nice and easy.
|33:19||here we are.
find the distance traveled in 5 seconds
|33:30||you love math?
so first thing youre going to want to do is youre going to want to factor that incredibly difficult quadratic t^2-4t+3 you factor that you get t-3 t-1=0 so far so good?
|34:00||you have a number line
and you should get that.
okay now if you want to find the distance this thing has traveled well theyre giving you 3 integrals the integral from 0 to 1 of t^2-4t+3 dt.
minus the integral from 1 to 3 of t^2-4t+3 dt
the integral from 3 to 4
of t^2-4t+3 dt.
howd we do on this part?
oh i said 5 seconds. sorry from 3 to 5.
so far so good?
if you break it up like this at least youll get some partial credit.
|35:02||alright lets do the antiderivative so we have
from 0 to 1.
minus... t^3/3 minus.. 2t^2 +3t
1 to 3
from 3 to 5.
so far so good?
we already get to this point?
so theyre all the same antiderivative so you could probably come up with more efficient ways to find the actual answer.
|36:02||so if you plug in you get
so thats 4/3 so for the first
traveled a distance of 4/3
now here you plug in and you get uhh 9... minus 18..
that gets you the +4/3.
and the third part is annoying 125/3 minus.. 50
you should get a total distance 28/3 units.
|37:31||alright thats enough.