Finding the Area Between Two Graphs

[XodoX]

Homework Statement

f(x)=x²-1 and f(x)=2x+2

Homework Equations

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.

 

[tiny-tim]

Hi XodoX!

So you integrate using those as upper and lower and plug it in and subtract, right?

yes

But I get 0 for each.

how?

show us your integrations​

 

[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.

 

[skiller]

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.

So at what point did you do any integration?

 

[tiny-tim]

but you haven't integrated!

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

go forth and integrate!

 

[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.

 

[Mark44]

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.

 

[XodoX]

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

 

[Mark44]

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.

 

[HallsofIvy]

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.

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