[MUSIC]. 
We've seen that we can use either 
horizontal strips or vertical strips when 
using integrals to calculate area. 
We end up computing the same number, the 
area of this region, whether we use 
vertical rectangles or horizontal 
rectangles, we're getting the same 
answer, the area of this region. 
But if both work, which one should we 
choose? 
The best choice depends on the shape of 
the region. 
I like to think about how many different 
kinds of edges my rectangles will touch. 
Here is an example to demonstrate what I 
mean. 
I've drawn some region in here with three 
edges, this curved edge, this curved 
edge, and this straight edge along the x 
axis. 
Now, I could calculate the area of the 
enclosed region. 
Either by decomposing this into vertical 
strips or horizontal strips. 
The bad news about vertical strips is 
that there's two different kinds of 
vertical strips. 
There's vertical strips over here, and 
there's vertical strips over here. 
The vertical strips over here touch the 
orange and the purple edge, and the 
vertical strips over here would touch the 
orange and the blue edge. 
Contrast that with the much better 
situation that happens if I cut this up 
into horizontal strips. 
All of my horizontal strips have an 
orange side and a blue side. 
I only have one type of horizontal strip. 
So I should calculate the area of a 
region like this using horizontal strips. 
Let's take this advice and apply it to 
solve a specific problem. 
So here's a bigger, specific version of 
this problem. 
Let's figure out the area of this region. 
this orange curve is the curve y equals 
the square root of x, and this blue curve 
is y equals the square root of 2x minus 
one. 
And then here I've just got the purple 
line along the x-axis. 
I've really got a choice as to how to 
approach this. 
I could do this with vertical strips. 
But then I'd have to do two different 
integrals. 
I'd have to handle the case over here 
where rectangles are touching the orange 
line and the purple line. 
And I'd have to do an integral over here 
where my rectangles are touching the 
orange line and the blue line. 
It's going to be better to use horizontal 
strips. 
In that case, all of the rectangles are 
the same kind of rectangle, they're a 
rectangle with one edge on this orange 
curve, and one edge on this blue curve. 
Let me just draw one of those thin 
horizontal rectangles. 
I have to figure out the size of that 
rectangle. 
Let's suppose this thin strip is at 
height y. 
I know how tall this strip is. 
I'll call that dy. 
But how wide is this green strip? 
I should figure out what this point and 
where this point is. 
And if I think about it a little bit, if 
this is y, I can solve this equation and 
find out that in that case x. 
Is y squared, and I can similarly solve 
this equation and find out that if this 
point has a y coordinate y, the x 
coordinate will be y squared plus 1 over 
2. 
And that tells me the width of this green 
rectangle, it's this minus this. 
I'm going to write down the area of that 
green rectangle. 
So the width is y squared plus 1 over 2 
minus y squared, that's how wide this 
rectangle is. 
And the height of that rectangle is dy, 
so this product is the area of that green 
rectangle. 
But of course I don't care just about 
that one rectangle, I want to integrate 
so that I'm adding up the areas of all 
kinds of thin rectangles. 
And then in the limit I'm getting the 
area of this region exactly. 
So instead of just this single rectangle, 
I'm going to integrate y goes from where 
to where. 
Well, I look at my picture here, y can be 
as small as zero and as tall as this 
point up here which is one. 
So y goes from zero to one and this 
interval calculates the area of that 
region. 
Let's do the calculation. 
This integral is the integral from 0 to 
1. 
for I can simplify the integrand a bit. 
It's 1 half minus y squared over 2 dy . 
And now we can write an antiderivative 
for that. 
So I add a derivative of 1 half. 
One half y and a derivative of y squared 
over 2 is y cubed over 6. 
I'm going to evaluate that at 0 and 1, 
take a difference. 
But when I plug in 0 I just get 0. 
So the answer is whatever I get when I 
plug in 1. 
And that's 1 half times 1, minus 1 cubed 
over 6. 
Could simplify that a bit, that's 1 half 
minus a 6th, and 1 half, well, that's 3 
6ths, so 3 6ths minus 1 6th, is 2 6ths, 
which you might write as, 1 3rd. 
So the area of that, shark fin shaped 
region is a third of a square unit. 
Often, it's not there are right ways or 
wrong ways but there are better ways and 
worse ways, right. 
Ways that are better in that they are 
quicker, safer, more elegant. 
So don't just make good choices. 
Make the best choices.