# Area between curves and integration

1. Jan 13, 2008

### Hurricane3

This isnt a homework question, but from my notes that I couldn't figure out (wasn't able to copy the rest of the down)

I have two functions:
y=x
y=x^3 , 0 $$\leq$$ x $$\leq$$ 2

and I have to find the area between these two curves.

I know how to do it with respect to x, but I have troubles integrating with respect to y.

From a quick sketch of the graph, it seems like I will have three different integrals.

I can formulate the first two definite integrals:

$$\int^{1}_{0}$$ (y^1/3 - y) dy + $$\int^{2}_{1}$$(y-y^1/3)dy

but I don't know how to forumate the last one... help!

The area should be 5/2.

Thanks

2. Jan 13, 2008

### Dick

I'm not sure why you inverted the problem into a dy integration. You can do that but then the upper limit is definitely not 2. Just integrate (x-x^3)dx between 0 and 1 and (x^3-x)dx between 1 and 2 and add them.

3. Jan 13, 2008

### Hurricane3

Because this was part of the notes. In the notes, we integrated with respect to x and the professor showed us how to integrate with respect to y as well. Except I didn't have time to copy that part of the notes (integration with respect to y), so I was just trying to do it just now, but I dunno whats the upper limit for the third part of the integration...

4. Jan 13, 2008

### Dick

If you need to do it that way, then y^(1/3)-y from 0 to 1. y-y^(1/3) from 1 to 2 and 2-y^(1/3) from 2 to 8. Do you see why? Draw the graph. x^3=y, 2^3=8.

5. Jan 14, 2008

### Hurricane3

yeah i see it now! thanks =)

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