Question

How do you find the area between `f(x)=sqrt(3x)+1, g(x)=x+1`?

Answer 1

The area is `3/2` square units.

Explanation:

Start by finding the intersection points.

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

`sqrt(3x) = x`

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

`3x = x^2`

`x^2 - 3x = 0`

`x(x - 3) = 0`

`x = 0 and 3`

These will be our bounds of integration. Next, we use a test point in `[0, 3]` to determine which function is higher up.

Let the test point be `x = 2`.

`f(2) = sqrt(6) + 1 ~~ 3.449`

`g(2) = 2 + 1 = 3`

Therefore, `f(x)` will be above `g(x)`. We can now find an expression for the area:

`A = int_0^3 f(x) - g(x)dx`

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

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

`A = int_0^3 sqrt(3x) - x dx`

`A = int_0^3 sqrt(3)sqrt(x) - xdx`

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

`A = 2/3sqrt(3)(3)^(3/2) - 1/2(3)^2 - (2/3sqrt(3)(0^(3/2) - 1/2(0)^2))`

`A = 6 - 9/2`

`A = 3/2" "u^2`

Hopefully this helps!