Your f(x)=sqrt[x+8]/(x+4)(x-6) is MUCH clearer, but not 100% clear. I will assume that your denominator here is the product of (x+4) and (x-6):
√x+8)
f(x) = ------------------
(x+4)(x-6)
We cannot divide by zero. Thus, x cannot be -4 and cannot be 6.
Thus, the domain is:
(-infinity, -4) or (-4, 6) or (6, infinity).
The
net area of the region in relation to the x-axis, is just the integral at those bounds, thus
![\bf \displaystyle \int\limits_{-\frac{\pi }{2}}^{\pi }~8cos(\theta )\cdot d\theta \implies 8\int\limits_{-\frac{\pi }{2}}^{\pi }~cos(\theta )\cdot d\theta \\\\\\ \left. 8sin(\theta )\cfrac{}{} \right]_{-\frac{\pi }{2}}^{\pi }\implies [8sin(\pi )]~-~\left[8sin\left(-\frac{\pi }{2} \right) \right]\implies [0]-[-8]\implies 8](https://tex.z-dn.net/?f=%5Cbf%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%5E%7B%5Cpi%20%7D~8cos%28%5Ctheta%20%29%5Ccdot%20d%5Ctheta%20%5Cimplies%208%5Cint%5Climits_%7B-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%5E%7B%5Cpi%20%7D~cos%28%5Ctheta%20%29%5Ccdot%20d%5Ctheta%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft.%208sin%28%5Ctheta%20%29%5Ccfrac%7B%7D%7B%7D%20%20%5Cright%5D_%7B-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%5E%7B%5Cpi%20%7D%5Cimplies%20%5B8sin%28%5Cpi%20%29%5D~-~%5Cleft%5B8sin%5Cleft%28-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%20%20%5Cright%5D%5Cimplies%20%5B0%5D-%5B-8%5D%5Cimplies%208)
Yes,
x^2 -4x+4
= x^2 -2x-2x +4
= x (x-2) -2 (x-2)
= (x-2)(x-2)
As you can see it has repeated factor.
Answer:
16 27/50
Step-by-step explanation:
