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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. Staff · 4 years, 3 months ago

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

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" ? · 4 years, 3 months ago

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. Staff · 4 years, 3 months ago

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 · 4 years, 3 months ago

This assumes that $$f(x)$$ and $$g(x)$$ are differentiable. Staff · 4 years, 3 months ago