# Homework Help: Defined integration general question

1. Jan 10, 2016

### Rumplestiltskin

Not a homework problem, so not sure if this is in the right area. Thought it was too sophomore to post in the maths forums.

So, when you integrate y within defined limits for x, you get the area between the graph and the x-axis within those limits for x. But
a) Why is this always between the graph and the x-axis, as if you've also defined limits for y (y=0 to y=maximum or minimum point)? Your answer would be infinite if you didn't define limits for y, so it's in there somewhere.
b) Wouldn't integration give you the area above and below the x-axis, within the defined limits? So in the below example, it would give you the area x + y. We had a problem in class where this parabola was transformed a little up, and when integrated only found the larger area x. Does integration only give you the larger area bounded by a graph and the x-axis?

2. Jan 10, 2016

### Staff: Mentor

Your graph is not very helpful, as you are using x and y to mean areas, rather than as labels on the horizontal and vertical axes.
Maybe I can clear up some of your misconceptions with an example. Let's assume that the graph you show is that of the parabola $x = y^2$, and we're interested in the area of the upper region (what you're calling x), bounded by the upper branch of the graph, the x-axis, and the line x = 1.

Using vertical strips of width $\Delta x$, the typical area elements have an area of $y \Delta x$. To get an integrand that we can intergrate, we have to have y as a function of x. From the equation $x = y^2$, we see that $y = +\sqrt{x}$, so our integral becomes $\int_{x = 0}^1 (\sqrt{x} - 0) dx.$

The reason I wrote $(\sqrt{x} - 0)$ was to emphasize that the upper end of each vertical strip is at $y = \sqrt{x}$ and the lower end is at $y = 0$, so the length of each strip is $\sqrt{x} - 0$.

If I wanted the area of the lower portion, the typical area element would be $(0 - (-\sqrt{x}))\Delta x$. The upper end of each strip is at y = 0; the lower end is at $y = -\sqrt{x}$. This time our integral would be $\int_{x = 0}^1 (0 - (-\sqrt{x})) dx = \int_{x = 0}^1 \sqrt{x} dx$. Clearly, this results in exactly the same as the integral for the upper portion, which should make sense.

3. Jan 10, 2016

### Rumplestiltskin

My bad, a and b.
That's a little too complex for me to digest at this point, I'm just a high school student who's thinking further ahead than he should be. Basic differential calculus is the most I'm acquainted with. I'd be happy to dive in the deep end to understand this, but could we start from basics? Vertical strips?

4. Jan 10, 2016

### Staff: Mentor

I'll try more direct answers to the questions you asked.
If you're interested in calculating the area between some graph and the x-axis, these are defined limits for y. In terms of your drawing (and assuming that it represents the graph of $x = y^2$), the area of the upper region would be infinity, as the region isn't bounded on the right. That's why I made the assumption that we wanted the area between the curve and the x-axis, and between x = 0 and x = 1.
In your first post you didn't define any limits that bound the region on the right. That's why I made the additional assumption that the area should be bounded on the right be the vertical line x = 1.

So the upper portion of the region is bounded on all four sides: on the left by x = 0 (a vertical line), above by the graph of $y = \sqrt{x}$, on the right by the line x = 1, and below by the x-axis (the line y = 0).

If the lower boundary were the lower branch of the curve instead of the x-axis, you would get an area that is twice the area of the upper portion.
I'm not sure what you're saying here without more details about the problem you worked.

5. Jan 10, 2016

### Ray Vickson

Actually, the integral of f(x) between x = a and x = b represents "area" only if f(x) ≥ 0 throughout the interval from x = a to x = b. If f(x) < 0 on the interval a → b then the integral is negative, and represents -Area, the negative of the actual area. If f(x) has both positive and negative values between x = a and x = b, the integral is the sum of all the areas above the x-axis minus the sum of all the areas below the x-axis.

This can cause problems if you are not careful. For example, suppose you plot the graph of y = f(x) on x from a to b, and you want to paint the areas between the graph and the x-axis. The amount of paint you will need equals the total area painted, times the amount of paint needed per unit of area. However, the integral itself could be zero. That does not mean you need 0 paint; it just means that you need to be careful when relating area to the integral when there are sign changes.

6. Jan 10, 2016

### Rumplestiltskin

Wasn't necessary as the graph made it clear that region a was bounded by a lower and upper limit, but sure, if it helps.
So how are limits for y defined? Why are you, by default, calculating the area between a curve and the x-axis when you integrate that curve?

This exactly. So who has defined the limits for y, y = 0 and y = f(x)? If I wanted to find the area between x = 0 to 1 and y = 1 to f(x), given that I couldn't simply reconsider my Cartesian axes, how would I go about doing that?
It's the y limits I'm interested in. My question was, how is y = 0 to y = f(x) presupposed in the integral? And why does integrating y = f(x) + z only give you the larger area bounded, when two areas are bounded between y = 0 and y = f(x) + z?

The graph y2 = x was moved up a little. So the area a bounded by the curve and the x-axis, was greater than the area b bounded by the curve and the x-axis.

Huh? a and b were the regions, not limits.

7. Jan 10, 2016

### Staff: Mentor

I wasn't sure that the vertical line you drew was a boundary for the region. You didn't identify that line, as a boundary. Also, when you said this in the first post --
-- that made me think that the vertical line was not a bound on the region. For a region to have a finite area, there need to be boundaries at top and bottom, as well as left and right.

For your second question above, I made assumptions about a possible problem description.
Those are usually stated in the problem description if they are looking for the area between the graph of y = f(x) and the x-axis.
$\int_{x = 0}^1 (f(x) - 1) dx$. Here I'm assuming that the graph of f lies above the line y = 1 throughout the interval [0, 1]. If that's not the case, things are a bit more complicated.
The integral $\int_{x = 0}^1 (f(x) - 0) dx = \int_{x = 0}^1 f(x)dx$ implies that the upper limit of the region is the graph of y = f(x) and the lower limit is the line y = 0 (the x-axis). The limits of integration imply that the left and right boundaries are the lines x = 0 and x = 1.
The area of the typical area element would be (<y-value at top of curve> - <y-value at bottom of curve>) * $\Delta x.$ Or you could integrate using horizontal strips.

8. Jan 10, 2016

### Ray Vickson

In your very first post you said " So, when you integrate y within defined limits for x, you get the area between the graph and the x-axis within those limits for x." I just called the defined limits 'a' and 'b'; if you want to call them something else, go ahead and do that, but it will not influence in any way the points I am making!

9. Jan 10, 2016

### Rumplestiltskin

But the vertical line is an defined limit of x, not y. What I meant by the infinite area was that, if you didn't have the (ostensibly) defined limits of y=0 and/or y = f(x), you'd have infinite area in that case.

Haha, sorry about the miscommunication - I'm not a very rigorous mind. By "who" I was being ironic and pointing out that, since the y limits aren't apparently defined, it looks like someone's just assumed them for their own intentions of finding the area between y = 0 and y = f(x).
What I mean to ask is, why are the y limits always taken to be y = 0 and y = f(x) when integrating? This must be something inherent to the integral.

That makes sense, but it still begs the above question.

Yes! Why? How?

I know. Imagine that in the graph I drew, the parabola was moved up a couple units in relation to the axes, such that the region x (we've since decided that a would be more appropriate) would be larger than area y (now b). If I integrated y = f(x), I'd get the area of the region a -- at least that's what we all got in class. Why? Why not also b, since that is also between y = 0 and y = f(x)?

10. Jan 10, 2016

### Rumplestiltskin

Chill, dude. Just thought you had them confused with the regions. Lemme take a second look:

The bolded answers my question b), thank you.

11. Jan 10, 2016

### Ray Vickson

I don't need to "chill", because I am not upset; maybe you are, but I don't know why. Nothing I said was insulting or demeaning in any way.

12. Jan 10, 2016

### Staff: Mentor

No, the vertical line is a boundary of the region you're integrating. I don't think that calling it a "defined limit of x" helps you understand things.
Well, if you didn't have those boundaries, you wouldn't have anything to integrate. In any case, if there was no right-hand bound on the region, the area between the graph of f and the x-axis would be infinite.
They aren't, necessarily. It depends on the problem statement and the region whose area you are supposed to find. If the problem asks you to "find the area between the x-axis and the graph of y = f(x), between x = a and x = b" then the integral will be $\int_{x = a}^b f(x)dx$. It's as simple as that.

If the problem asks you to "find the area between y = 2 and y = f(x), between x = a and x = b" then the integral will be $\int_{x = a}^b (f(x) - 2)dx$.
I don't see how. I've explained it as clearly as I can.
You can read it off directly from the first integral I wrote, $\int_{x = 0}^1 (f(x) - 0) dx$
The interval over which integration is done is [0, 1], represented by the limits of integration.
Typical area element is $(f(x) - 0)\Delta x$, which represents the y-coordinate at the top of the area element - the y-coordinate at the bottom of the strip. The typical area element is almost identical to the integrand, with the only difference being $\Delta x$ in the typ. area element vs. dx in the integral.
[/quote]You're oversimplifying here. Because of the orientation of the graph (a parabola opening to the right), you don't just simply "integrate y = f(x)."

To quantify things a bit, let's suppose that your graph is the equation $(y - 2)^2 = x$. This does NOT represent y as a function of x because for each x > 0, there are two y values.

What is region A? (Let's not use either x or a to label the region.) Is it the region inside the parabola and above the x-axis? That does not define a region with finite area. To make a region with a finite area you need a boundary on the right.

I think this problem is well beyond your current abilities, so I think the best thing is for you to hold off until you know more about setting up integrals.

13. Jan 10, 2016

### Rumplestiltskin

I'm not sure you answered the question on second thoughts. Here's the problem I had in mind:

I had to find the area of region a. I did this by integrating the parabola within limits x = 0 and x = 1 and taking away triangle b. This worked. But if as you say this gave me the area above minus the area below, that would negate a pretty large chunk of the above area, which would imply that this wouldn't work out.

It's a limit of x and hence a boundary of the region. That's how I understand it.

I'm aware of this. I said that if you didn't have a bottommost boundary, your area would be infinite, and you seem to be in agreement. That was a statement, not a question.

Thanks. I'm going to read between the lines here and presume that those limits are in fact inherent to the integral. A PM I received seems to confirm this, so I'll do some more reading into the nature of integration.

14. Jan 10, 2016

### Staff: Mentor

What do you mean by "integrating the parabola"? I'm very curious to see the integral you set up.

Also, if your sketch is reasonably accurate, region "b" is not a triangle. It has three straight sides, and a short curved fourth edge. The figure is roughly triangular in shape, so approximating the area by the area of a triangle will be close but not exact.

15. Jan 12, 2016

### Rumplestiltskin

Sorry I haven't been able to respond.
I integrated y = √x + c (don't remember exact problem but take c as 1) within the limits x = 0 and x = 1.

My bad - region b was definitely a triangle. The parabola may have been a little to the left (its turning point < x = 0) to allow for this.
So not only that was funny, but just integrating y = √x within limits x = 0 and x = 1 gives you an area of 2/3. Area below has not been taken away from area above.