Integral

Integral step

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Integral step

So your area for this function, or for this integral, is going to be 1.

I'm telling you that this delta of x is a function such that its integral is 1.

If we're taking the integral from 0 to infinity of this thing, we already said what does this integral look like or what does this function look like?

The Laplace transform of any function is equal to the integral from 0 to infinity of that function times e to the minus st, dt.

Remember, this definite integral is really just the area under this curve of this whole function, of the unit step function times all of this stuff.

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

And that's taking the integral.

So the Laplace transform of some function of t is equal to the improper integral from 0 to infinity of e to the minus st times our function.

It'll actually be the integral from 4 and a half to 5 and a half of this probability density function or of this probably density function, the x.

I'm going to tell you that if I were to take the integral of this function from minus infinity to infinity, so essentially over the entire real number line, if I take the integral of this function, I'm defining it to be equal to 1.

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?

So it's kind of borrowing that notation, because this function of s is kind of an integral of y of t.

And then over that boundary, the function is a constant, 1 over 2 tau, so we could just take this integral.

The Laplace transform of some function f of t is equal to the integral from 0 to infinity, of e to the minus st, times our function, f of t dt.

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.

The integral of u prime times v.

And this is a definite integral, right?

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?

I'll just use that old integral sign.

What happens if I take the integral?

Let me write a nicer integral sign.

Because the t's are involved in evaluating the boundaries, since we're doing our definite integral or improper integral.

So if we have this unit step function, this thing is going to zero out this entire integral before we get to c.

So we can rewrite this integral as y.

So what's this integral going to look like?

Generally speaking, this integral cannot be solved analytically.

They're an integral part of the penile skeleton.

I give you an integral factor of 0.004.

How do I take the integral of that?

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