This also helps us make sense of the area of a triangle.

You're probably used to thinking of it as half the area of the bounding rectangle

by using geometry and similar triangles. But we could do things a little bit

differently. If we also arrange this triangle so that

the base is along the X axis and shear it horizontally.

So that it becomes a right triangle. We see, we have not changed the area

since we've preserved the horizontal area elements.

Now the hypotenuse has the equation of y equals h over b times x.

Now let's consider a slightly different area element.

This one, veritical The height of that vertical strip is, from our equation for

y, h/b*x, the thickness is dx, and thus, computing the area is the intergral of

the area element, and what do we get. We get the integral of h over b x dx as x

goes from 0 to b. Now, what do we see?

Well, h over b is a constant x integrated step x squared over 2.

Evaluating from 0 to b gives h b squared over 2b.

Minus 0. Simplifying we get one half base times

height. Very simple.

The example of a circular disc finaly takes us to a nonlinear case.

The classical means for determining his area was to use an Angular variable.

And to approximate as a limit of triangles.

Let us do likewise. This is perhaps not exactly how

Archimedes did it, but it's kind of close.

Consider an angular variable, theta. And consider a wedge.

Defined by an infinitesmal angle d theta. Then, thinking of this wedge as

approximately a triangle. Not exactly, but approximately a triangle

to first order, what would the dimensions be?

Well, we would say that the height of the triangle is roughly r, the radius of the

circle. The base is going to have length r times

d theta. With that in mind, the area element is

what? It is one half the base times the height,

that is r times r d theta. The area, being the integral of the area

element, is the integral of 1/2 r squared d theta, as theta sweeps all the way

around from zero to 2 pi. Now, we know what, well, wait a minute,

there's no theta in the integrand. There's just r squared, which is a

constant, as is 1/2, so we can pull out the 1/2 r squared.

The integral of d theta is, of course, theta.

Evaluating that from zero to 2 pi gives 2 pi times 1/2 r squared.

That is, pi R squared, the classical answer done very simply.

But this is not the only way to do this problem.

Consider if we used a radial variable, and we sweep out thin annuli and fill up

the disc in that manner. Let's call that radial variable t.

The annular strip is going to have thickness dt, and now we need to argue

that the area of this infinitesimal annular strip is The circumference, 2 pi

t times the thickness, d t. I'm not going to argue that that's really

correct right now. Let me just state it, and let's see what

happens when we integrate it. We get the integral of 2 pi t d t Aa t

goes from zero to r, the radius. That is, we get pi t squared as t goes

from zero to r. That again yields our answer, which, if

nothing else, tells us that our area element was indeed correct.

That is a very simple integral. But notice that it's not the only way

that we can do it. Still, we can cut the disk into slices.

Let's say vertical slices. And we sweep from left to right.

Placing the circular disk at the center of the x,y plane.

Allows us to use a strip of width dx, and height given by what?

Well, the equation of the circle is x squared plus y squared equals r squared.

Solving for y gives plus or minus square root of r squared minus x squared.

The area element, in this case, is going to be the height, that is, twice root r

squared minus x squared, times the width dx.

Then we see that integrating to get the area gives the integral of twice root r

squared minus x squared dx. Evaluating the limits from minus r to r,

what do we get? Well, this integral is not as easy as the

other two that we've done. It is an even integrand or a symmetric

domain, so could pull out the two, double it, and integrate from zero to r.

But we're still not done. We would need to use trigonometric

substitution, something of the form x equals r sine u and do a little bit of

work. I'm going to leave it to you to verify

that what one gets for that integral is pi r squared over 4.

Leading to the final answer that we all know and love.

Notice that depending on which way you split the domain up into area elements.

The resulting integral may be trivial, or may be quite involved.

This then leads to the example that we all know and love, the area between two

curves. Let's say f on the top, and g on the

bottom. Now, it's very clear given what we've

done, how to determine the area element. In this case, using a vertical strip,

it's height is going to be f(x) minus g(x).

its width is d(x). And so, integrating this, we see the

classical formula that you've no doubt seen before.

The area is the integral f or x minus g of x dx, as x goes from a to b.

One interesting application of this formula is to economics.

In the definition of something called the genie index.

This is a ratio that is used to quantify income and equality.

In a fixed population, it relies on one slightly technical definition and the

rest is simple calculus. >> Let us denote by f of x, the

fraction of the total income earned by the lowest x fraction of the populous,

the domain is x goes from 0 to 1 and you should think of x as a percentage of the

population And f(x) as the amount of income earned by that lowest x percent.