Answer to Question #261348 in Calculus for Artika

Question #261348

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

Expert's answer

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"

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