00:01
Let's evaluate the definite integral.
00:04
Here we have a fraction in the intigrant.
00:07
Let's go ahead and do partial fraction to composition here.
00:11
The first thing we should check is if this quadratic factors, and it does.
00:17
This is t minus one.
00:19
Let's do t plus one squared and then t minus one squared.
00:26
So this tells us that the partial fraction should look like a over t plus one, plus b over t plus 1 squared.
00:37
This is what the textbook calls case 2, repeated linear factors, c over t minus 1, d over t minus 1 squared.
00:52
Now let's go ahead and multiply both sides of this equation over here by this denominator in the left.
01:00
When we do so we have 1, then we have a, t plus 1, t minus 1 squared.
01:11
For b we have t minus 1 squared.
01:13
For b we have t minus 1 squared.
01:17
For see we have t minus 1 then t plus 1 squared and for d we just have t plus 1 squared so on the right hand side let's just go ahead and multiply out and simplify as much as we can so go ahead and foil this or multiply out this quadratic and then multiply to t plus 1 then the coefficient we have a and then we have a cubic after simplifying.
01:53
Then for b we just have go ahead and foil that.
01:59
For c we have some work to do here.
02:01
This is t cubed, t squared minus t minus 1.
02:08
And finally for d t squared two t and a 1.
02:16
Now on the right hand side, let's go ahead and factor out a t cubed.
02:22
We have a plus c then let's pull out t squared.
02:28
Have minus a plus b plus c plus d and then pull out a t minus a minus 2b minus c plus 2d and then finally the constant term a plus b minus c plus d now the left hand side this is all equal to one if we want we could write the left hand side is zero t cubed 0 t squared plus 0t plus 1 and then we see that we must have a plus c equal 0 because those are the coefficients of t cubed similarly the coefficient in front of t squared those have to be equal the coefficients in front of t have to be equal so we have another equation this is all equal to zero and then we have the constant term on the left is 1 on the right hand side there's our constant term so those must be equal and this gives us a 4 by 4 system in the coefficients a through d so let's write out these equations and solve the system on the next page we had a plus c equals zero minus a plus b plus c plus d equals zero minus minus a minus 2b minus c plus 2d equals 0 minus 2 b minus c plus 2d equals 0 and then a plus b minus c plus d equals 1.
04:36
From this first equation we can go ahead and solve that.
04:45
A equals minus c if we want to go ahead and solve for a.
04:49
And then we can go ahead and plug this into each of the equations for a.
04:54
So when we plug it into the second equation, that becomes a c.
04:59
So we have b plus 2c plus d equals 2c...