Find the area of the region enclosed by the graphs of y= 3/x and y=4-x
Given: "y=3\/x ,\\ y=4-x"
Point of intersection of these two curve will occur at
"3\/x=4-x\\Rightarrow x^2-4x+3=0\n\\\\\\Rightarrow (x-3)(x-1)=0\n\\\\\\Rightarrow x=1,3"
Graph of the above function is shown below
Area of the region enclosed by these two curves
"A=\\int_{1}^{3} (4-x-\\frac{3}{x})dx=[4x-\\frac{x^2}{2}-3lnx]_{1}^{3}\\\\\n=12-\\frac{9}{2}-3ln3-4+\\frac{1}{2}+3ln1=4-3ln3" ["\\because ln 1=0" ]
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