And that's where the integral comes in.

If I took the integral from minus infinity to infinity of d sub tau of t dt, what is this going to be equal to?

You could write just d dot w there, but we could say, we'll do a line integral along this curve c, could call that c or along r, whatever you want to say it, of dw.

So once again, that number represents the area under the curve here, this area under the curve.

And we're going to see some concrete examples of taking a

And if you watched all of the videos on integration and the definite integral and area and a curve, you realize that an integral is essentially a sum, it's kind of an infinite sum of a bunch of these infinitely small dy's.

Integral

So the equation of this is x squared plus y squared is equal to 1 has a radius of 1 unit circle.

So what the cumulative distribution function is essentially let me call it the cumulative distribution function it's a function of x.

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

Integral step

Let's see.

You've already dealt with vectors.

I'll call the plus the constant due to y.

Show integral

Welcome back.

That's equal to that.

That's the intuition behind the definite integral.

That'll come in useful later on.

Now, this is the part, and we've done this before, it's a

So let's say that this is x right here, that's our x.

Gaussian density has a wider width.

Line segments

Segment

segment

But for some constant y, what if I could just figure out the area under the curve there?

And the a's cancel out.

Almost every Laplace transform problem turns into an integration by parts problem.

Vectors Segments

Segment Axis

So how do we evaluate this integral?

Where f of x,y is equal to P of x, y i plus Q of x, y j.

I'm now going to use definite integrals to figure out the areas under a bunch of curves and, if we have time, maybe even between some curves.

And this is all going to be really good integration practice for us.

So that's minus 1 over s e to the minus st, times v, sine of at, minus the integral.

The perpendicular line through a given segment's mid point.

You'll get the area of this rectangle, which might be a pretty good approximation for the area under the curve.

Now, if we take the integral of both sides of this equation, we get uv is equal to the antiderivative of u prime v plus the antiderivative of uv prime.

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

Well, now we have another hairy integral we need to solve.

If that's the case, then what is v?

So let's do it traditionally.

So let's do x just to show you it really doesn't matter.

The bottom boundary is this surface.

And we have its derivative sitting right there.