[MUSIC]. 
Even integrating something as "easy" as a 
polynomial can be rather surprising. 
Let's integrate from x equals 0 to 1 just 
the polynomial x plus 10 times x minus 1 
to the 10th power dx. 
So there's some integration problem, just 
an integration problem for a polynomial. 
Your first inclination might be oh, I 
know what to do, just have no fear. 
I'll just expand everything in sight. 
So, we've got x plus 10, times x minus 1, 
to the 10th. 
Which you could expand by, you know the 
binomial theorem if you want. 
We're going to get x to the 11th minus 55 
x to the 9th. 
Plus 330 x to the 8th minus 990 x to the 
7th plus 1,848 times x to the 6th mi, I 
mean this is ridiculous. 
Arr, yeah, so I mean yeah, you, you could 
do that. 
You could just expand everything out. 
But there's a better way right. 
We can make use of U-substitution. 
Yeah, we're not going to do that, let's 
do something else. 
So instead let's set u equal x minus 1, 
and I could solve this for x. 
That means that x is u plus 1. 
All right, if I just add 1 to both sides. 
And I could make this substitution in 
this problem. 
So this integral from x goes from 0 to 1, 
is, what's x plus 10 if x is u plus 1. 
I could write u plus 1 plus 10. 
And whats x minus one, thats just u to 
the tenth power, and whats dx well if u 
is x du is dx. 
So its the same as trying to do this 
integration problem. 
In some text books this technique of 
startign with the use of substitution but 
then solving for x. 
This is sometimes called back 
substitution. 
regardless of the name, let's proceed. 
So this is the integral, x goes from 0 to 
1 of u plus 11 times u to the 10th du. 
I could expand this, and that's the same 
as the 
Integral x goes from zero to one of u to 
the 11th. 
Plus 11 u to the 10th du. 
But this is just asking me to anti 
differentiate a polynomial which I can 
definitely do. 
Alright. 
This is u to the 12 over 12. 
Plus, what's an anti-derivative of 11 u 
to the 10th, it's just u to the 11th. 
But now I've got a little bit of an 
issue, right? 
I've got x going from 0 to 1. 
So here is u going from what to what? 
Yeah, so how do we deal with those end 
points? 
Well, when x is 0, u is minus 1. 
So I'll put a minus 1 there. 
And when x is 1, u is 0. 
So I'll put a zero there. 
And now, this is just as easy as plugging 
in zero, plugging in minus 1, and taking 
the difference. 
So what do I get when I plug in zero? 
I just get zero. 
And what do I get when I plug in minus 1? 
Well, I get minus 1 to the 12th over 12. 
Plus minus 1 to the 11th. 
Alright minus 1 to the 12 is just 1. 
So this is 1 12th. 
And minus 1 to the 11th well that's 
negative 1. 
So I've got 0 minus what's this a 12 
minus one that's negative 11th 12th. 
But at 0 minus negative 11, 12ths. 
So the integral is 11, 12ths. 
So the calculation turned out to be 
pretty nice. 
I mean 11 12ths, isn't a particularly 
difficult fraction to come up with. 
The crazy thing to think about is that we 
could have approached this just by 
expanding everything in sight. 
Yeah, I really could have expanded this 
thing out, and then just started 
anti-differentiating you know. 
And finding this really was you know, I 
don't know x to the 12th over 12, minus I 
think it's 11 halves, x to the tenth plus 
a hundered and ten thirds x to the ninth 
right. 
And it would keep on goign and id just 
evaluate that, exit one exit zero and 
take the differenc i mean i really could 
have done this by expanding. 
And somehow that would of ended up giving 
me the same answer right, of eleven 
twelths. 
Except that had I gone down that path, I 
would have been certain to have walked 
right into an arithmetic error. 
The world of mathematical solutions is 
not divided into correct solutions and 
incorrect solutions. 
It's really more subtle than that. 
Mathematical solutions come in a ton of 
varieties. 
Right, there's easy solutions and hard 
solutions. 
There's solutions that are defending you 
against arithmetic errors. 
And then there's solutions that are just 
walking you right into an arithmetic 
mess. 
The fun of u substitution isn't just 
being able to do integrals, it's being 
able to do integrals in an efficient way 
that protects you from those arithmetic 
errors.