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If f(x) and g(x) are both continue fungtions such that f(g(x))=x and g(f(x) )=x, f(0)=0,f(3)=29. find ∫_0^3▒f(x)dx + ∫_0^29▒g(x)dx.
Find the area of a region bounded by the curve y=1/(1+x^2 ) , x positive axis and y positive axis.
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TopNewestNote that Nidhin makes the assumption that \(f(x), g(x) \) are differentiable functions. This is not a necessary assumption (not is it valid, since we can have kinks).

Hint:The direct approach is to observe that \(f(x) \) and \(g(x) \) are inverse functions of each other. Since \(f(x), g(x) \) and are continuous functions, this implies that \(f(x), g(x) \) are strictly increasing functions.Log in to reply

thanks,,, thats very kind of you... :)

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thanks nidhin,, btw i'm a vocational school student and no so smart in calculus :) do you know how can i see the solutions to the previous week problems in "challenges" ?

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Hi Robi,

In the weekly challenges box, there is a tab that says "solutions" every wednesday(thursday depending where you are in the world) the solutions to all of the previous weeks problems that you looked at will be revealed.

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integral 0 to 29 g(x)dx= integral 0 to 3 g(f(t))f'(t)dt,

so total integral becomes integral 0 to 3 ( f(t)dt + g(f(t)f'(t)dt) but g(f(t))= t,

so integral = integral 0 to 3 (f(t) + tf'(t)dt )= integral d(tf(t))dt= [tf(t)] from 0 to 3= 87

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This assumes that \(f(x)\) and \(g(x) \) are differentiable.

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