# Can Someone check my work?

1. Nov 27, 2005

### kenny87

I am to find the area of the region between the graphs of y=x^2 and y=-x from x=0 to x=2

I evaluated the integral from 0 to 2 of x^2 - -x
I got 14/3

Is this correct?

2. Nov 27, 2005

### georgeh

are you sure the function is y=-x?

3. Nov 27, 2005

### kenny87

yeah, i just double checked them. y=x^2 and y= -x

4. Nov 27, 2005

### georgeh

if you draw the two functions, there is no area to integrate between those curves for those limits. I am kind confused why they would say from x=0 to x=2.
I setup the problem, and evaluated the integral with respect to y
and got 1/6.

i said, integral from y=1 to y = 0 and y^(.5)-y dy

5. Nov 27, 2005

### kenny87

I think that they're going for the area beneath x^2 and above -x with the boundaries of x=0 and x=2.

I know 1/6 isnt the correct answer because its not one of the choices...
the choices are

2/3, 8/3, 4, 14/3, and 16/3

these are ap problems and i know they try to trick you though, so im really not sure if my answer is correct or not... i might have made up some math

6. Nov 27, 2005

### georgeh

ooooh, i am sorry.. yeah then your integral ends up being x^3/3 - x^2/2
and when you put the limits in the answer should be 2/3.
(2)^3/3-2^2/2 = 2/3

7. Nov 27, 2005

### kenny87

can you explain to me the process you used to get that?

what i did:

1/3*x^3 - 1/2*x^2 and then evaluated it at 2 and 0.... what did i do wrong?

8. Nov 27, 2005

### georgeh

arr, stupid mistake..
lets call f(x)=x^2 and g(x)=-x
then integral from f(x)-g(x) = x^2+x
integrate that function
and we get x^3/+x^2/2
plug in the limits
and we get 14/3..
sorry for the wrong answer sheesh.. i feel dumb.

9. Nov 27, 2005

### kenny87

no problem. thank you so much!