# Area under a curve question

1. Feb 6, 2013

### lionely

1. The problem statement, all variables and given/known data
Find the area bounded by the curve y= x2-x-2 and the x-axis
from x=-2 and x=3.

The attempt at a solution

I integrated from x=-2 to x=3 using (x^3/3)-(x^2/2)-2x and I got -4 5/6
but the answer is -4 1/2 .

I don't really see where I went wrong.

2. Feb 6, 2013

3. Feb 6, 2013

### SteamKing

Staff Emeritus
Good News! You're both wrong.

Did you try checking your math again?

4. Feb 6, 2013

### lionely

I still get - 4 5/6 :S

5. Feb 6, 2013

### SteamKing

Staff Emeritus
What I originally meant was the book answer is also wrong.

Remember, the integral is F(3) - F(-2)

6. Feb 6, 2013

### lionely

When I do that I get -4 5/6!!
((3)^3/3)- (9/2)- 2(3) - [(-2)^3/3) - (2) - 2(-2)]

7. Feb 6, 2013

### Karnage1993

You guys are all wrong. Look carefully that he said it's also bounded by the x-axis. You guys are counting the NEGATIVE area under the x-axis which you are not supposed to. The area is

$\displaystyle\int\limits_{-2}^{-1} x^2 - x - 2\ dx + \int\limits_{1}^{3} x^2 - x - 2\ dx = 5/2$

8. Feb 6, 2013

### lionely

5/2? really? The answer in the book is that far off?

9. Feb 6, 2013

### Karnage1993

Sorry, a mistake in the limits. It's actually:

$\displaystyle\int\limits_{-2}^{-1} x^2 - x - 2\ dx + \int\limits_{2}^{3} x^2 - x - 2\ dx = 11/3$

See here for a better understanding: http://i.imgur.com/U8iWBwJ.png

10. Feb 6, 2013

### skiller

I don't. How is the answer not -5/6 ?

11. Feb 6, 2013

### lionely

but shouldn't the 2nd integral be from -2 to 3?

12. Feb 6, 2013

### Karnage1993

Why is that? You don't want to count the negative area, do you?

13. Feb 6, 2013

### lionely

Oh I see what you did.. so the answer in my book is 100% wrong. so it's 11/3? oh and may I ask what program you used to produce that sketch?

14. Feb 6, 2013

### Karnage1993

Actually, thinking about it some more, I realized that you would count the negative area. It says bounded by the curve and x axis and not above the x axis. So yes, the area would be -5/6.

That particular diagram was from Mathematica 8, but I'm sure you can make one using WolframAlpha.

15. Feb 6, 2013

### haruspex

It's one of those awkward 'areas'. I was always taught to treat area under the x-axis as negative. This gives the right result in most practical circumstances, e.g. distance advanced when y=velocity, x=time.
If you want to count all areas as positive then it becomes $\int_{-2}^3|x^2-x-2|dx$. I see no basis for simply ignoring the area below the axis. May well have gone wrong in the arithmetic somewhere, but I make that 49/6.

16. Feb 6, 2013

### lionely

:S so confused like 3 different answers.

17. Feb 6, 2013

### haruspex

Yes, -5/6, thanks.

18. Feb 7, 2013

### lionely

I'm confused I don't see how you guys got -5/6 =/

19. Feb 7, 2013

### skiller

The area bounded by the curve and the x-axis, if you count below as negative, is -5/6 by a simple integral from -2 to 3.

If you count the area below the x-axis as positive too, then the answer is 49/6, by integrating between -2 to -1, -1 to 2, and 2 to 3, summing the absolute values of each.

If you discount any area below the x-axis altogether (bizarre if you ask me), then you just integrate between -2 to -1 and 2 to 3, giving 11/3.

20. Feb 7, 2013

### HallsofIvy

That's an impossible answer. Area is always positive.
From x= -2 to x= -1, the graph is above the x-axis. The integral is positive so that area is just the integral from -2 to -1. From x= -1 to x= 2, the graph is below the x-axis. The integral is negative so that area is the negative of the integral from -1 to 2. Finally, from x= 2 to x= 3, the graph is above the x-axis. The integral is positive so the area is equal to the integral from 2 to 3.

The area bounded by the x-axis, the graph of $y= x^2- x- 2$, and the vertical lines x= -2 and x= 3, is given by
$$\int_{-2}^{-1} x^2- x- 2 dx- \int_{-1}^{2} x^2- x- 2 dx+ \int_{2}^{3} x^2- x- 2 dx$$

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