If f : R → R , f ( x ) = x + sin x , then find the value of
∫ 0 π f − 1 ( x ) d x
Note : f − 1 ( x ) denotes the inverse function of f ( x ) .
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this can also be seen from the perspective of the area of the other part of the square with side 'pi' whose given part's area is integral of the given function.
Doing it by the help of graph was easier
l e t x = t + sin t , d x = ( 1 + cos t ) d t ⇒ f − 1 ( x ) = f − 1 ( t + sin t ) = t & x = 0 ⇒ t = 0 x = π ⇒ t = π ∫ 0 π f − 1 ( x ) d x = ∫ 0 π t ( 1 + cos t ) d t = [ 2 t 2 + t sin t + cos t ] 0 π = 2 . 9 3 4 8
I used the following method. First, i found the area enclosed by f(x) an then subtracted the area enclosed by the line y=x. Then, i subtracted the found area twice from that of f(x) to get the answer. :) Of course your method is better.
Given that f ( x ) = y = x + sin x and to find I = ∫ 0 π f − 1 ( x ) d x . Swapping x with y , then we have x = y + sin y and to find I = ∫ a b y d x . And from x = y + sin y ⟹ 0 = a + sin a ⟹ a = 0 , similarly b = π .
x ⟹ d x ⟹ ∫ 0 π x d x ∫ 0 π y d x = y + sin y = d y + cos y d y = ( 1 + cos y ) d y = ∫ 0 π y d x + ∫ 0 π sin y d x = ∫ 0 π x d x − ∫ 0 π sin y d x = 2 π 2 − ∫ 0 π sin y ( 1 + cos y ) d y = 2 π 2 + cos y ∣ ∣ ∣ ∣ 0 π − ∫ 0 π 2 sin 2 y d y = 2 π 2 − 2 + 4 cos 2 y ∣ ∣ ∣ ∣ 0 π = 2 π 2 − 2 + 0 ≈ 2 . 9 3 5
the general property is
∫ a b f ( x ) + ∫ f ( a ) f ( b ) f − 1 ( x ) = b f ( b ) − a f ( a )
Looking at the graph of f (x), can it be said that the inverse is same as f (x)?
Nope! f − 1 ( x ) = f ( x ) here
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If a , b are fixed points of f ( x ) and f ( x ) is monotonic and differentiable, then ∫ a b [ f ( x ) + f − 1 ( x ) ] d x = b 2 − a 2 Since f ( 0 ) = 0 and f ( π ) = π , then ∫ 0 π f − 1 ( x ) d x = π 2 − ∫ 0 π ( x + sin x ) d x = 2 π 2 − 2 ≈ 2 . 9 3 4 8
Note that, this technique won the 2013 MIT Integration Bee .