Indefinite integrals (part II)

Examples of taking the indefinite integral (or anti-derivative) of polynomials.

Welcome back.

In this presentation I'm just going to do a bunch of examples

of taking the antiderivative or the indefinite integral of

polynomial expressions, and hopefully I'll show you that

it's a pretty straightforward thing to do.

So let's get started.

If I wanted to take indefinite integral and

you could do a web search for integral and

you'll see this drawn properly-- take the indefinite integral--

let me make a big expression.

Let's say I want to take the indefinite integral of

3x to the -5 - 7x to the third + 3- x to the ninth.

So you might already be intimidated by what I wrote down.

Well, one, if you saw the last presentation

or if you understood presentation, you probably realize,

well the indefinite integral

even though it looks like fancy math, isn't that fancy.

Or at least it isn't that difficult to perform.

And all you have to realize now is

if we took the derivative of a polynomial,

which was just the sum of the derivatives of each of the terms.

that actually turns out is the same way the other way around.

The antiderivative of this entire expression is just

the sum of the antiderivatives of each of the individual terms.

So we can just take the antiderivatives of each term

and we'll get the answer.

So what does this equal?

Well in this case 3x to the minus 5 power.

So we take the exponent, we add 1 to the exponent,

so now we get x to the negative 4,

and then we multiply the coefficient times 1 over the new exponent.

So 1 over the new exponent is minus 1/4.

So 3 times minus 1/4 is minus 3/4.

And let's see. Here we have x to the third.

So instead of x to the third, let's raise it by one number.

So we get x to the fourth.

And then we multiply the coefficient.

You know, we could either just keep the minus

and say the coefficient's 7,

or we could just say the coefficient is minus 7.

We multiply the coefficient times 1 over the new exponent.

So the new exponent is 4,

so we multiply 1/4 times minus 7, so minus 7/4.

And now this is interesting.

3, just 3.

Well how do we apply this?

Well isn't 3 the same thing as 3 times x to the 0?

Right, because x to the 0 is just 1.

And that's how you should view it.

It shows you that this rule is actually very consistent.

So what's the antiderivative of 3?

Well if we view 3 as 3 x to the 0, we raise the exponent by 1,

so now we're going to have x to the 1.

And x to the 1 is just x,

so I'm just going to leave it as an x.

And we multiply it, the old coefficient-- this 3

or you know the derivative coefficient--

we multiply that times 1 over the inverse of the new exponent.

So the exponent's 1, so the inverse of 1 is 1,

so it just stays 3.

We've multiplied 3 times 1/1, which is still just 3.

And then finally x to the ninth

--I think you're getting the hang of this--

we raise the exponent by one, x to the tenth.

And then we multiply the current coefficient.

Well the current coefficient is minus 1, right.

We just didn't write the 1 there.

We multiply the current coefficient minus 1 times

1 over the new exponent, so it's minus 1/10.

There we did it.

That wasn't too difficult of taking the antiderivative

or--once again I always forget.

Plus c, right?

Because when you take the derivative of any constant

it becomes 0, so it might have disappeared here.

So plus c where this is any constant.

This could be a 10, could be a million,

could be a minus trillion.

It's any constant.

And just to really hit the point home,

let's take the derivative of this

and just make sure we got this expression.

And hopefully this is second nature to you by now.

And you know if you ever run out of practice problems

in your book because you love doing math so much,

just make up problems.

That's what I'm doing.

I do this even when I'm not recording videos, just for fun.

So let's take the derivative of this.

Minus 4 times this coefficient.

Minus 4 times minus 3/4 is 3x.

Then we subtract 1 from this exponent, minus 5.

And then minus 7/4 x 4 is -7

-- we take 1 from this exponent-- x to the third.

And I promise you I'm not even looking up here.

I know you might think, well Sal, he's just looking up here,

but no I'm actually in my head at least working through this.

And then plus the derivative of 3x.

Well the derivative of 3x is 3-- is almost second nature

now, but you can kind of do this-- is 3x to the 1.

And you say 1 times 3 is 3 times x to the 0.

And then 10 times minus 1/10.

Well that's just minus 1.

x to the 1 less than 10, so x to the ninth, plus--

what's the derivative of any constant?

Right, it's 0.

You could almost do this constant

as some number times x to the 0.

And if you took the derivative,

well you multiply the 0 times c and you get 0.

Well, you might get minus 1 depending on how you're doing it.

But that's actually kind of an interesting question.

OK I'll stop digressing.

But you get a 0 here, and if you simplify that,

that just equals 3x to the -5 - 7x to the third +3- x to the ninth.

Think we have time for one more problem like this.

I think you probably got this.

This is probably one of the more straightforward things

you'll learn in mathematics.

And in future presentations

I'll give you more of an intuition of why the antiderivative is useful.

We're learning the indefinite integral,

but we could learn use the definite integral,

which we'll learn in a couple of presentations

to figure out things like the area under a curve,

or the volume of a rotational figure.

Well I don't confuse you too much.

Let's do one more problem.

I won't make this one as hairy.

So the integral of -1/2x to the -3 plus 7x to the fifth.

Let's start with this term of the polynomial.

We raise the exponent one, so x to the minus 2 now, right,

because we added one to negative 3.

And then we multiply 1 over this new exponent

times the old coefficient.

And actually I'll write out all the steps.

So the old coefficient is minus 1/2.

So this is a minus 2.

Minus 2 so we multiply it times minus 1/2.

Let me switch colors back.

Plus we raise the exponent by one, x to the sixth,

and we multiply the old coefficient times

1 over the new coefficient, times 1/6.

And so what's the answer?

Well what's minus 1/2 times minus 1/2?

Well that's positive 1/4 x to the minus 2.

Oh, and of course, plus c.

As you can tell, this is my main source of

missing points on calculus quizzes.

1/4 x to the minus 2 plus 7/6 x to the sixth plus c.

There you go.

And if you wanted to take the derivative, minus 2 times 1/4

is minus 2/4 which is minus 1/2 x to the minus 3.

And then 6 times 7/6 is 7x.

And then you decrease the exponent by one,

x to the fifth.

And the derivative of our constant is 0.

And then we get our original expression.

Hopefully at this point you're pretty comfortable

taking a derivative of a polynomial,

and then given a polynomial you can actually take the
antiderivative,

go the other way.

And never forget to do your plus c.

And I hope you understand why we have to put that constant there,

because when you take an antiderivative,

you don't know whether the original thing that you took the derivative of,

I guess, had a constant there, because the constant's derivative is 0.

Hopefully I confused you with that last statement.

I'll see you in the next presentation

and I'll show you how to reverse the chain rule.

See you soon.