Use a double integral to find the area of the region whose upperbound is the function y = (x+3)^1/2 the lower boundary is y = (x+3)/3
I got 4.5
find the point of the intersection
$\displaystyle \sqrt{x+3}=\frac{x+3}{3} $
$\displaystyle 9(x+3) = x^2 + 6x+9 $
$\displaystyle x^2 -3x-18 = 0 $
$\displaystyle (x-6)(x+3) = 0 \Rightarrow x=6,x=-3 $
$\displaystyle \int_{6}^{-3} \int_{\frac{x+3}{3}}^{\sqrt{x+3}} dy dx $
I got 4.5 to