# Calculator program gives incorrect results (Definite Integrals/ Area)

1. Dec 16, 2007

### lLovePhysics

Calculator program gives incorrect results (Definite Integrals/ Area)????

I've inserted this definite integral into my calculator program:

$$\int_{-1}^{2}x(x^2-4)dx$$

and my calculator gives me -9/4 for the integral, which is what my book's answer key has written down.

However, the area that it gives me is 23/4. Is this the correct area or what? Why are the definite integral and area different? Aren't they essentially the same thing?... or did my calculator program fail on me?

(It worked for just about every integral except this one. That is, the integral and area values matched.)

Thank you very much.

Last edited: Dec 16, 2007
2. Dec 16, 2007

### rocomath

i just compured -9/4 and my calculator also shows the same, so it's just a typo.

3. Dec 16, 2007

### lLovePhysics

Wait, I don't quite understand what you mean. What's a typo? Is my calculator wrong?

My Calculator gives me -9/4 for the integral and 23/4 for the area.

My book gives me the same integral.

However, I'm just curious as to why the definite integral and the area are different. Aren't they suppose to be the same? Aren't definite integrals and Riemann Sums equivalent?

Last edited: Dec 16, 2007
4. Dec 16, 2007

### rocomath

the answer for the definite integral is -9/4

how did you compute the integral? let me re-work it just incase cuz i just thought of something.

Last edited: Dec 16, 2007
5. Dec 16, 2007

### lLovePhysics

Also, when I graph the equation f(x)=x(x^2-4) on my graphing calculator and look at the area bounded by [-1,2], f(x) and the x-axis, it seems to me that the answer should be postive thus making -9/4 an incorrect answer for the area. Why is the definite integral different from the area bounded by the same f(x) region and upper/lower limits?

Might it be because you need to divide up the definite integral since there is a negative area and a postive area?

EDIT: (Wait... I think that -9/4 should be the answer for the area because the larger area is below the x-axis.... Is my calculator program flawed then??) Or I don't know, what is the right answer here for the area??

Last edited: Dec 16, 2007
6. Dec 16, 2007

### rocomath

That's right, it has to be positive. Idk what I was thinking.

Anyways, after graphing it. A1 is from [-1,0] which is positive, and A2 is [0,2] which is negative.

so A1-A2 = 23/4, which is what you got.

If I'm wrong again, plz correct me! I need to review Calculus 1, I feel so dumb! ha :p

7. Dec 16, 2007

### Feldoh

$$\int x(x^2-4)dx$$

$$u = x^2-4$$
$$du = 2x dx$$

$$\frac{1}{2} \int u du = \frac{(x^2-4)^2}{4}$$

$$\frac{(x^2-4)^2}{4}|_{-1}^{2} = \frac{-(-3)^2}{4} = \frac{-9}{4}$$

8. Dec 16, 2007

### lLovePhysics

Ohh.. I found the explanation in my book. After thinking, area cannot be negative! heh, what was I thinking?

So I guess there are cases when the definite integral cannot equal the area bounded by the function and the upper/lower limits. Am I correct?

9. Dec 16, 2007

### Feldoh

No, the definite integral will give you the area bounded by the lower and upper bounds of integration

In this case x(x^2-4) is an odd function with zero's at 4,-4,0. Since the function is odd then the area bounded from -2 to 2 would be zero where [-2,0] is the positive portion and [0,2] is the negative portion. It makes sense that since we're only going from [-1,0] in the positive that there will indeed be more negative.

The integral calculates the change of a function between two intervals given the said functions derivative function, it CAN represent the area under a curve but we can consider it to be negative because it's not really area as much as a change of a function, it's just convention to call everything below the x-axis negative.

Last edited: Dec 16, 2007
10. Dec 16, 2007

I do not think you are correct. And depending on what the question is asking you, area can certainly be negative. Anything that goes below the x-axis is negative area.

The definite integral of sinx from 0 to 2pi is 0. However, the total area is not. That is to say: Signed Area=0 and Net Unsigned Area= positive number (<---this is the physical area bounded by curve)

Casey

11. Dec 16, 2007

### lLovePhysics

My textbook says, "For a definite integral to be interpreted as an area (as defined by Riemann Sums) , the function f must be continuous and nonnegative on [a,b]. "

Are you guys saying something other than the physical area? For example, Ft=J (Impulse)?

12. Dec 16, 2007

Physical Area must be positive.
Area under curve need not be, For example $Work=\int f(x) dx$ can be negative.

13. Dec 16, 2007

### Feldoh

f(x) = x(x^2-4)
f(1) = 1(1^2-4) = -3

As I said earlier it's just convention to call everything under the x-axis negative, however the area is certainly NOT negative in a PHYSICAL SENSE. The concept is a bit abstract but that's the normal convention...

Does that make sense?

14. Dec 16, 2007

### lLovePhysics

Gotcha! Thanks guys

15. Dec 16, 2007

### lLovePhysics

I think I get what you mean: It is like the net area (that represents something else like Work = Fd) Right?

16. Dec 16, 2007

Think about it. What does the definite integral do. Picture a simple horizontal line at say y=5. What does it mean to take the definite integral of y=5 from 0 to 5?

As a Riemann sum it should click immediately. It is a super-thin, vertically standing rectangle whose area is being added to the next rectangle and so on.....

Now if the x-axis represents time in seconds, and the y-axis represents velocity in m/s, than the 'area' being base times height represents a quantity whose units are
s*(m/s)=m. This is displacement.

Now if the velocity curve at some point went below the x-axis, that would imply that we reversed directions somewhere, hence velocity is negative, thus some of my displacement is too.

But the actual physical area...the amount of space bounded by the curve is positive. Physical area cannot be negative. It would not make sense to say the field is -25 square meters!

Physical area--->positive
The 'stuff' it represents-->not necessarily.

Casey

17. Dec 16, 2007

### lLovePhysics

Thanks a lot for making it all clear Casey!