# `y = sqrt(x), y = (1/2)x, x = 9` Sketch the region enclosed by the given curves and find its area.

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You need to determine first the points of intersection between curves `y = sqrt x` and `y = (1/2)x` , by solving the equation, such that:

`sqrt x = (1/2)x => x = x^2/4 => x^2/4 - x = 0`

Factoring out x yields:

`x(x/4 - 1) = 0 => x = 0 orx/4 - 1 = 0`

Hence, the endpoints of integral are x = 0 and x = 4.

You need to decide what curve is greater than the other on the interval [0,4]. You need to notice that`(1/2)x < sqrt x` on the interval [0,4], hence, you may evaluate the area of the region enclosed by the given curves, such that:

`int_a^b (f(x) - g(x))dx` , where `f(x) > g(x)` for `x in [a,b]`

`int_0^4 (sqrt x - x/2)dx = int_0^4 sqrt x dx - (1/2)int_0^4 x dx`

`int_0^4 (sqrt x - x/2)dx = ((2/3)x*sqrt x - x^2/4)|_0^4`

`int_0^4 (sqrt x - x/2)dx = (2/3)(4sqrt 4 - 0*sqrt 0) - (4^2/4 - 0^2/4)`

`int_0^4 (sqrt x - x/2)dx = 16/3 - 4`

`int_0^4 (sqrt x - x/2)dx = (16-12)/3`

`int_0^4 (sqrt x - x/2)dx = 4/3`

Hence, evaluating the area of the region enclosed by the given curves, yields `int_0^4 (sqrt x - x/2)dx = 4/3.`

The area of the region enclosed by the given curves is found between the red and black curves, for ` ` `x in [0,4].`

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