Answer to Question #265103 in Calculus for seed

Question

Find the area of the region enclosed by the graphs of y= 3/x and y=4-x

Answer 1

Given: $y=3/x ,\ y=4-x$

Point of intersection of these two curve will occur at

$3/x=4-x\Rightarrow x^2-4x+3=0 \\\Rightarrow (x-3)(x-1)=0 \\\Rightarrow x=1,3$

Graph of the above function is shown below

Area of the region enclosed by these two curves

$A=\int_{1}^{3} (4-x-\frac{3}{x})dx=[4x-\frac{x^2}{2}-3lnx]_{1}^{3}\\ =12-\frac{9}{2}-3ln3-4+\frac{1}{2}+3ln1=4-3ln3$ [$\because ln 1=0$ ]