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April 18, 2014

April 18, 2014

Posted by **Pat** on Friday, October 4, 2013 at 2:28am.

y=x, y=2√x

- Calculus -
**Jai**, Friday, October 4, 2013 at 3:56amReq'd: area bounded by y = x and y = 2*sqrt(x)

First thing to do here is to find their points of intersection, so we'll know the bounds. We can do it algebraically or graphically.

To find algebraically the points of intersection, we just use substitution. Since y = x,

y = 2*sqrt(x)

x = 2*sqrt(x)

x^2 = 4x

x(x - 4) = 0

x = 0 and x = 4

To set up the integral, we choose whether vertical strips (dx) or horizontal strips (dy) to be used. In here, let's just use vertical (dx). The bounds for this is from x = 0 to x = 4, and since it is dx, all expressions must be in terms of x.

Therefore,

Integral (2*sqrt(x) - x)dx from 0 to 4

= (2*(2/3)*(x^3/2) - (1/2)x^2) from 0 to 4

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

= [ ((4/3)(4^(3/2)) - (1/2)*4^2 ] - [ ((4/3)(0^(3/2)) - (1/2)*0^2 ]

= (4/3)(8) - (1/2)(16) - 0

= 32/3 - 8

= 32/3 - 24/3

= 8/3 square units

Hope this helps~ :3

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