Find the area of the region bounded by the graphs f(x)=sqrt(3x)+1, g(x)=x+1

Did you make a sketch?

first find their intersection.
I had them intersect at (0,1) and (3,4)

Area = [integral] ((3x)^(1/2) + 1 - x - 1) from 0 to 3
= ((2/9(3x)^(3/2) - x^2/2) from 0 to 2
= (2/9)(9)^(3/2) - 9/2 - (0-0))
= (2/9)(27) - 9/2
= 6 - 9/2 = 3/2

To find the area of the region bounded by the graphs of the given functions, you need to determine the points where the graphs intersect first. The area can then be calculated by taking the definite integral of the difference between the two functions over the interval between these points.

Let's find the intersection points:

f(x) = g(x)
sqrt(3x) + 1 = x + 1

Simplify the equation:

sqrt(3x) = x

To solve this equation, square both sides:

3x = x^2

Rearrange to bring all terms to one side:

x^2 - 3x = 0

Factor out x:

x(x - 3) = 0

This equation gives us two possible values for x: x = 0 and x = 3.

Now, we can calculate the area of the region:

∫[a,b] (f(x) - g(x))dx

where [a, b] is the interval between the intersection points.

For our case, the interval is [0, 3]. So, the area can be calculated as:

Area = ∫[0, 3] (sqrt(3x) + 1 - (x + 1))dx

Simplifying the integral:

Area = ∫[0, 3] (sqrt(3x) - x)dx

Now, you can evaluate this integral to find the area of the bounded region by antiderivatives or integration techniques.