# Area between two curves

1. Aug 27, 2008

### duki

I have three questions:

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

Find the areas of the regions whose boundaries are given.

2. Relevant equations

$$y=x^3-3$$
$$y=1$$

3. The attempt at a solution

x=-2, x=2
I got -10.67 but I know this can't be true because you can't have a negative area.

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

Find the areas of the regions whose boundaries are given.

2. Relevant equations

$$y^2=x$$
$$x+y=2$$

3. The attempt at a solution

y=1, y=-2
I got -4.5, but again that can't be right because it's negative :(

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

Find the areas of the regions whose boundaries are given.

2. Relevant equations

$$y=x^3-2x^2-3x$$
$$y=0$$

3. The attempt at a solution

I got to x=0, x=-1, and x=3 but I don't know where to go from here.

Thanks for any help! :)

2. Aug 27, 2008

### Dick

Let's just start with the first one. How did you get x=-2, x=2 or was that given? The curves y=x^3-3 and y=1 don't enclose any bounded region.

3. Aug 27, 2008

### duki

many apologies... it should have been x^2-3

4. Aug 28, 2008

### HallsofIvy

Staff Emeritus
If you get a negative area then you have the two functions in the wrong order. For x between -2 and 2, the graph of x2- 3 is below the graph of y= 1. You should be integrating $\int [1- (x^2-3)]dx= \int (4- x^2)dx$. That, integrated between -2 and 2, is positive.

5. Aug 28, 2008

### moemoney

And for the last problem, you would have two different enclosed areas. One would be between -1 and 0 and the other between 0 and 3.
This means you would have to set up 2 different equations and find the sum of the areas.
HINT: The equations are very similiar just siwtched in order.