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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.