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

##### 1 Answer

The area is

#### 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

Let the test point be

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

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

Therefore,

#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!