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integration

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.

Note by Robi Fajar Bahari
4 years, 3 months ago

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Note 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. Calvin Lin Staff · 4 years, 3 months ago

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thanks,,, thats very kind of you... :) Robi Fajar Bahari · 4 years, 3 months ago

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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" ? Robi Fajar Bahari · 4 years, 3 months ago

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@Robi Fajar Bahari 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. Peter Taylor Staff · 4 years, 3 months ago

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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 Nidhin Kurian · 4 years, 3 months ago

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@Nidhin Kurian This assumes that \(f(x)\) and \(g(x) \) are differentiable. Calvin Lin Staff · 4 years, 3 months ago

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