# Area under curve

1. Oct 22, 2008

### uradnky

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

Find the area bound by y= x - 2$$\sqrt{x}$$ and y=0

3. The attempt at a solution

X intercept = 4
Y intercept = 0

Using verticle rectangles..

dA = ((Upper curve)-(Lower curve)) dx
$$\int$$(dA = $$\int$$(0-(x-2$$\sqrt{x}$$)) dx

A= -(1/2X^2)-((2x^(3/2))/(2/3)) Evaluated from 0 to 4

A= 16

I understand this is an extremely easy question but I just cannot see where I have gone wrong with it. Is there a problem with my signs since the area I am finding is below the x-axis? Sorry if this is a bit messy, any help is appecricated, thanks.

2. Oct 22, 2008

### steelphantom

You should be calculating $$\int$$2sqrt(x) - x from 0 to 4.

3. Oct 23, 2008

### uradnky

That was easy. Thanks.

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