Answer to Question #261348 in Calculus for Artika

Question

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

Answer 1

area:

$A=\int^3_1(4-x)dx-\int^3_1(3/x)dx=(4x-x^2/2-3lnx)|^3_1=$

$=12-4.5-3ln3-3.5=4-3ln3=0.704$