Question

How do you sketch the region enclosed by `y=1+sqrtx, Y=(3+x)/3` and find the area?

Answer 1

`4.5` square units

Explanation:

Start by finding the intersection points of the two curves.

`1 + sqrt(x) = (3 + x)/3`

`3(1+ sqrt(x)) = 3 + x`

`3 + 3sqrt(x) =3 + x`

`3sqrt(x) = x+3 - 3`

`(3sqrtx)^2 = x^2`

`9x = x^2`

`0 = x^2-9x`

`0 = x(x-9)`

`x = 0 and 9`

Next, sketch a rudimentary graph of the curves to see which one lies above the other. You will find the graph of `y = sqrt(x) + 1` is above `y = (x + 3)/3`. You can trace the square root function by drawing the function `y = sqrt(x)` and transform it one unit up. `y = 1/3(x + 3)` is equivalent to `y = 1/3x + 1`, which is your run of the mill linear function.

Integrate:

`=int_0^9 sqrt(x) + 1 - (x + 3)/3 dx`

`=int_0^9 (3sqrtx + 3 - x - 3)/3dx`

`=int_0^9 (3sqrtx - x)/3 dx`

`=1/3int_0^9 3sqrtx -xdx`

`=1/3[2x^(3/2) - 1/2x^2]_0^9`

`=1/3(13.5)`

`=4.5` square units

Hopefully this helps