Integral

So what is our integral going to become with this substitution?

Integral step

Show integral

Integral step

And what we're concerned with is the line integral over this curve c.

And this is a definite integral, right?

Well, the first integral I'm integrating with respect to x.

And now I take the integral with respect to y.

I'll keep the improper integral with us the whole time.

So what's the integral of this expression with respect to y?

So if we take from zero to infinity, what I'm saying is taking this integral is equivalent to taking this integral.

And when you do the integration by parts, this could be an indefinite integral, an improper integral, a definite integral, whatever.

Just with the helmet, integral helmet, it's really no problem to breathe.

So what's the integral of 1 over y with respect to y?

The integral of 1 is y, the integral of y squared, well that's minus y to the third over 3.

And that's taking the integral.

This is equal to the integral from 0 to infinity.

But what we'll do is we'll define the triple integral.

And this is equal to the integral over the region.

We said, this integral is just the original problem again.

That this integral is equal to the double integral over the region this would be the region under question in this example.

So the integration by parts just tells us that the integral of uv prime is equal to uv minus the integral of

Set here the spatial integral step.

Anyway, integral from 0 to infinity.

Minus this integral evaluated at 0.

That's what that double integral represents.

And what's the integral of this?

I'm going to tell you what the integral of this is.

We took the integral with respect to t. e to the minus sA, right?

So we have this integral of u prime v is going to be equal to this, uv minus the integral of uv prime.

Let's now do another triple integral, and in this one I won't actually evaluate the triple integral.

So anyway, let's take the anti derivative and evaluate this improper definite integral, or this improper integral.

Well, we could factor out an e to the minus sc and bring it outside of the integral, because this has nothing to do with what we're taking the integral with respect to.

This is essentially just this is the indefinite integral of x squared plus 1.

I could do it by anything because this is a definite integral.

And then that is equal to the integral from 0 to infinity.

The Laplace Transform for our purposes is defined as the improper integral.

The integral of u prime times v.

And that's where the integral comes in.

So what's the integral of this thing?

Let's try to set up this integral.

That's the intuition behind the definite integral.

Let's just worry about the indefinite integral.

So how do we evaluate this integral?