00:01
Okay, so we want to evaluate this improper integral and improper integral is whenever it has that infinity in its bound okay, so what can we do here? well, what we can do here is do a u -sub so i'm gonna let u to be negative x cubed we can take the derivative move to 3 to the front and take 1 from the power and so we would get so x squared dx we have in it.
00:25
So why don't we divide both sides by negative 3? what would happen is we get du over negative 3 equals x squared dx so we can replace that with du negative 3 negative x cube can be replaced with u now be perfect we can take out this 5 out of the integrand integral so we'll get 1 over infinity x squared dx is du over negative 3 and each negative excuse me to you we can pull out this negative 3 on the bottom and this will give us negative 5 thirds divide of integral 1 to infinity each to you du now we take the antiderivative e to you that would just give us e to the u and we plug in our u back for x we'll get e to the negative x cube.
01:08
So that's what we're gonna do right now so antiderivative e to u is e to u and we plug in the u back for x and be e to negative x cubed so keep in mind that since we're dealing with limit we're dealing with 1 to infinity.
01:23
You can't actually evaluate at infinity so actually before we move on let's rewrite it we can rewrite as negative 5 thirds change this into a limit replace the infinity with t and we say 1 to t e to the u du and then we can do what we just said earlier and then take the antiderivative so i'm to do if you use e to you plug in our u back in terms of x so be e to negative x cube evaluated at 1 and t so we have a negative 5 third on the outside and we're still taking the limit as t approaches infinity now we're gonna do f of b minus f of a so i'm gonna put t into our x's and 1 into our x's so that would give us negative 5 third limit as t approaches infinity of e to negative t cube minus e to the negative 1 cubed notice that t approaches infinity.
02:20
There's only one t and we would apply it there.
02:23
So to give us some space i'm gonna erase the top and continue on the top.
02:28
So we're gonna have 9 5 third limit t approaches infinity.
02:33
Let's rewrite these e to the negative t third same as 1 over e to the t cubed so you bring it to the bottom it becomes positive exponent minus e to the notice that negative 1 cube well 1 cube is 1 and then you have a negative 1 and likewise we can rewrite that as 1 over e to the positive 1 and that's what we get if we do the limit as t approaches infinity here, i'm gonna think about it logically e to the infinity cube well, i'll just round side e to the infinity cube infinity cube is like infinity and e is a constant 2 .71 to infinity.
03:12
It's just gonna get bigger and bigger and bigger and bigger so in other words as you plug in values for t and as they get closer to infinity your denominator will get bigger and bigger and bigger if your denominator gets bigger and bigger and bigger your overall value will get smaller and smaller just like how if i had 1 half and then i move the denominator to 1 tenth to 1 one hundredth to 1 1 millionth as we go down we're getting smaller and smaller as these values on the bottom get bigger so they keep getting smaller and smaller...