# Area bounded between 2 graphs

1. Jul 23, 2012

### XodoX

1. The problem statement, all variables and given/known data

f(x)=x2-1 and f(x)=2x+2

2. Relevant equations

3. The attempt at a solution

Points of intersection are -1 and 3. So you integrate using those as upper and lower and plug it in and subtract, right? But I get 0 for each. So nothing to subtract and 0 is not the correct answer.

2. Jul 23, 2012

### tiny-tim

Hi XodoX!
yes
how?

3. Jul 23, 2012

### XodoX

Well, you set both to 0 and basically combine them, then you get x^2 - 2x - 3. If you plug in -1 and 3 for it, then you get 0.

4. Jul 23, 2012

### skiller

So at what point did you do any integration?

5. Jul 23, 2012

### tiny-tim

but you haven't integrated!!

all you've done is find the points where their difference is 0​

go forth and integrate!!

6. Jul 23, 2012

### XodoX

Never-mind. Wrong number.

I get -10.6 after integrating the combined equation. Plug in 3 and subtract it from what I get for -1.

x3/3 - x2-3x

No, it's +10.6. Sorry.

7. Jul 23, 2012

### Staff: Mentor

An area should be nonnegative. If you got a negative number, your integrand is set up incorrectly.

Note that the line is above the parabola throughout the interval.

8. Jul 23, 2012

### XodoX

Yes, I did. I did the 3 first and then the -1. I thought that's how you did that.

9. Jul 23, 2012

### Staff: Mentor

I'm talking about the integrand, not the limits of integration. I believe you set up the integral as:
$$\int_{-1}^3 (x^2 - 1) - (2x + 2)~dx$$

That will give you a negative number.

10. Jul 24, 2012

### HallsofIvy

Staff Emeritus
Because you have them "wrong way around". $2x+2> x^2- 1$ for all x between -1 and 3.