calculus, u-substitution for definite integrals
i i feel that there must be a way but i don't know and yeah it's very difficult i think
in fact i don't even know if this is possible or not thanks to skillshare for sponsoring this video
okay quick question for you guys which area is bigger a or b
all right answering three two one they are actually the same kind of surprising right
let's have a look right here for this region first it's a weird shape right but we do know that we
have a function and it goes from one over e to e so to find this area we can just integrate it and
the idea of the integral is that we use rectangles so let me just draw one right here for you guys
denote the base right here is dx because that's just a small change in terms of the x value
and then the vertical distance is just a y and then to find the area of this rectangle
is just base times height which is y times dx and let's call this to be the differential area
like so and then to find the whole thing we'll just have to add up all of such rectangles
and to add in calculus the superversion is to integrate all right and when we integrate all the
differential areas we get the area and then notice that we have y in the x world which is not allowed
so we are going to write y as this which is square root of 1 minus parenthesis ln x squared all
over x and then we have the dx here and we will have to denote that it goes from 1 over e to e
and now this right here is a pretty scary rookie calculus one integral right but it's not so bad
because we can just do the circle u substitution let's go ahead and u equal to the inner function
that's usually a good idea so let's see and then once we put on standard substitution we are going
to differentiate both sides we get du that's equal to the derivative of l and x is 1 over x
dx and we also need to get the dx by itself so let's multiply the x on both sides
all right now we will take this integral to the new world and the idea is the following
originally the picture is based on the x-axis and also this is the equation and you see that
the region the shape is just not something that we are familiar with once we do this let's take
a look notice that x goes from 1 over e we will also have to change that to u to 1 over e to here
we are looking at u equals ln of 1 over e and that will give us negative 1.
so we are going to get the integral in terms of u it will start with negative 1 and then put in here
u is going to be ln of e which is 1. so u goes from negative 1 to 1 and then we have the square
root 1 minus ln x is the u and then we square that over here this x and then the dx is this
xdu this part is extremely important it tells you which word we are in not only that notice once we
replace the dx with x d u the x here and the x here cancel very nice because you have to keep
in mind that x is not allowed in the u world so what exactly are we going to get uh in terms of u
we are looking at the integral let me just write everything read like this
u goes from negative 1 to 1 and then we have square root of 1 minus u squared d u
this integral is equivalent to that integral in terms of their values so you know what
let's try to graph this right now 1 minus u squared in the square root it looks like this
and they will be equal to that it's just going to give us a semi-circle just the top
half like this and then it goes from negative 1 to 1. and this right here is the u axis and uh
you know what the equation is technically y equals square root of one minus u squared
and do we know how to find the area of a semicircle yes we do this right here is just
one half of pi times the radius which is one so we can just do one and we square that work
that out we get pi over 2. so you see after the substitution if we pick the right one if we are
able to change a weird shape into something that we know much better hey we might not even need to
do any crazy integration technique to figure out the area of this shape we did
pretty nice but now i would like to give you guys another integral that's also equal to pi over 2.
let's have a look right here notice if we integrate from negative 1 to 1 of 1 over 1 plus
and let me just use another variable that say t and we square that and then dt let me first
tell you that if we graph this we will get a picture like this this is how one over t squared
looks like and we are also going from negative one to 1 and this is on the t axis the t world
i would like to tell you this area is also equal to pi over 2. all right so the other same size and
let me just convince you guys this real quick because this is not like a shape that we know
integrate this we will get inverse tangent of x and then we're just going to be plugging numbers
all right first one we get pi over four nothing first tangent of one and then minus
inverse tangent of negative one we get negative pi over four which is just going to be a plus
so everything together we do get pi over 2 which is this value here
so in terms of areas they are equal but now here is the question that
i tried many times and have no idea how to do it or maybe i don't even know if it's possible
how is it possible to do a substitution so it can go from here to here well maybe
if you would like you can go from here back there so my question for you guys is what sub
do we need in order for us to do this are we going to get t is equal to something in terms of u
right so this right here will be just in terms of u or if you prefer you can also try u is equal to
expression in terms of t i i feel that there must be a way but i don't know and
yeah it's very difficult i think why in fact i don't even know if this is possible or not
so hopefully somebody can give me an answer that if it's not possible tell me it's not
possible and let me know why if it's possible then let me know the substitution and what's up
uh that we are supposed to do with this all right hey just real quick have you noticed the
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