Why Is My Integration of the Region Between y=2x and y=x^2+3x-6 Incorrect?

[FuturEngineer]

Homework Statement

Find the region bounded by y= 2x and y = x^2 + 3x - 6.
I found the points of intersection to be x= -3, 2 by setting the equations equal to each other and solving for x.
I concluded that y = x^2+3x-6 is bigger since I tried a point in between the points of intersection and it came out to be greater.

Homework Equations

y =2x
y= x^2+3x-6
x= -3, 2

The Attempt at a Solution

I tried integrating x^2+3x-6 dx from -3 to 2. But it doesn't work. What am I doing wrong?
I also tried separating the integrals and integrating from -3 to 0, and from 0 to 2, but doesn't seem to work either. The back of my book says that the answer should be 81/32 but I don't know how they got there. Help!
Thanks!

 

[FuturEngineer]

Forgot to mention it should be integrating x^2+3x-6 - 2x (the other line).

 

[SammyS]

Forgot to mention it should be integrating x^2+3x-6 - 2x (the other line).

Does this mean that now you do know what to integrate ?

 

[LCKurtz]

Homework Statement

Find the region bounded by y= 2x and y = x^2 + 3x - 6.
I found the points of intersection to be x= -3, 2 by setting the equations equal to each other and solving for x.
I concluded that y = x^2+3x-6 is bigger since I tried a point in between the points of intersection and it came out to be greater.

Did you try $x=0$?

 

[FuturEngineer]

Did you try $x=0$?

No, why 0 though? Then y= 2x == 0 and the other equation would be -6. Is that what you mean?

 

[FuturEngineer]

Does this mean that now you do know what to integrate ?

That's what I tried integrating but its not correct according to my book...

 

[LCKurtz]

Forgot to mention it should be integrating x^2+3x-6 - 2x (the other line).

Did you try $x=0$?

No, why 0 though? Then y= 2x == 0 and the other equation would be -6. Is that what you mean?

The reason I suggested trying $x=0$ is it is the easiest, hence less error-prone, number to try. It might have prevented whatever error you made. You have your upper and lower curves reversed.