Find the area (in green) of the region bounded by the curves $f(x) =\sqrt x$, $g(x)=x-6$ and the x-axis by integral with respect to x
Attempt:
since $x-6 =\sqrt x$
$x=9$
$A = \displaystyle\int _a^b [f(x)-g(x)] dx$
$A = \displaystyle\int _0^9 \sqrt x -\int _0^9x-6$
$=\frac{2}{3}(9)^{\frac{3}{2}}-\left(\frac{1}{2}(9)^2-6(9)\right) $
$=31.5$ $unit^2$
I don't think this method is correct since I am not getting the same area when I solve the integral with respect to y which is also different when I take the whole area under f(x) then subtract the area of the triangle on the right (base=3, height=3)
Can anyone help?
Best Answer
Unfortunately, your integral grabs too much region. It's more like this:
To get just the area of the green region, you would have to use two separate integrals. $$ A = \int_0^6 \sqrt{x}\,dx + \int_6^9\left(\sqrt{x} - (x-6)\right)\,dx $$
Alternatively, you can integrate with respect to $y$. The right edge can be expressed as $x=y+6$, the left as $x=y^2$, between $y=0$ and $y=3$. So the area is $$ A = \int_0^3 (y+6 - y^2)\,dy $$ I hope that helps.