91Ó°ÊÓ

To find the points of intersection of the two curves, we need to set the two functions equal to each other and solve for \(x\): \(2-|2-x| = \frac{3}{|x|}\) Since there are absolute value expressions on both sides of the equation, we should consider the four possible combinations of positive and negative signs for the absolute values to find all solutions. Case 1: \((2-x)>0\) and \(x>0\) Solving for \(x\) in this case: \(2-(2-x) = \frac{3}{x} \Rightarrow x=3\) Case 2: \((2-x)>0\) and \(x<0\) Solving for \(x\) in this case: \(2-(2-x) = \frac{3}{-x} \Rightarrow -x^2=3x-6\) No real solutions for this case. Case 3: \((2-x)<0\) and \(x>0\) Solving for \(x\) in this case: \(2-(x-2) = \frac{3}{x} \Rightarrow x^2-3x+2=0 \Rightarrow x=1,2\) Case 4: \((2-x)<0\) and \(x<0\) Solving for \(x\) in this case: \(2-(x-2) = \frac{3}{-x} \Rightarrow -x^2-3x+6=0\) No real solutions for this case. Our points of intersection are: \((1,1), (2,0),\) and \((3,1)\).

We can sketch the given functions on the coordinate plane to get a better understanding of the regions they enclose: • For the function \(y=2-|2-x|\), we have a V-shaped figure with a vertex at \((2,0)\), a left-side slope of 1, and a right-side slope of -1. • For the function \(y=\frac{3}{|x|}\), we have two branches of a hyperbola, one on the right of the \(y\)-axis, the other on the left of the \(y\)-axis. Plotting these functions and their points of intersection, we find that the region enclosed by them is a triangle with vertices at \((1,1), (2,0),\) and \((3,1)\).

Since the enclosed region is a triangle, we can integrate the difference of the functions over the interval \([1,3]\) and take the absolute value to find the area: \(A=\int_{1}^{3}[(2-|2-x|) - (\frac{3}{|x|})]dx\) We need to split this integral into two parts since the functions have different definitions on the interval: \(A = [\int_{1}^{2}[(2-(2-x)) - (\frac{3}{x})]dx + \int_{2}^{3}[(2-(x-2)) - (\frac{3}{x})]dx]\) Now we can go ahead and compute these integrals: \(A = [\int_{1}^{2}(x-3-\frac{3}{x})dx + \int_{2}^{3}(-x+5-\frac{3}{x})dx]\) Solving the integrals, we obtain: \(A = [|(-\frac{1}{2}x^2+3x+3\ln|x|) \Big|_1^2 + (\frac{1}{2}x^2+5x-3\ln|x|) \Big|_2^3]\) \(A = |(-\frac{3}{2}+3-3\ln 2) - (-\frac{1}{2}+3-3\ln 1) + (\frac{9}{2}+15-9\ln 3) - (\frac{4}{2}+10-6\ln 2)|\) \(A = |(-\frac{1}{2}+3\ln3-8\ln2+3)|\) Computing this result, we get a final numerical answer for the area: \(A \approx 1.987\) So, the area enclosed by the given curves is approximately 1.987 square units.