Question

Let r be the region bounded by the graphs of `y= sqrt(x)` and `y=x/2`, how do you find the area of R?

Answer 1

The area of R `=10/3`

Explanation:

Let's find the intercepts of the two graphs.
`sqrtx=x/2` `=>` `x=x^2/4` `=>` `x(x/4-1)`
so, `x=0` and `x=4`
a small area is `dA=(sqrtx-x/4)dx`
`:. A=int_0^(4)(sqrtx-x/4)dx`
`=[2/3x^(3/2)-x^2/8]_0^4=16/3-2=10/3`
graph{(y-sqrtx)(y-x/2)=0 [-4.19, 6.907, -2.38, 3.167]}