Inverse Area with definite integration.!

Calculus Level 4

If f : R R f: \mathbb {R \to R} , f ( x ) = x + sin x f(x)=x+\sin x , then find the value of

0 π f 1 ( x ) d x \large \int_{0}^{\pi} f^{-1}(x) \ dx

Note : f 1 ( x ) f^{-1}(x) denotes the inverse function of f ( x ) f(x) .

Try more Integral problems here .


The answer is 2.935.

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5 solutions

If a , b a,b are fixed points of f ( x ) f(x) and f ( x ) f(x) is monotonic and differentiable, then a b [ f ( x ) + f 1 ( x ) ] d x = b 2 a 2 \int_a^b \left[ f(x) + f^{-1}(x) \right] \, dx = b^2 - a^2 Since f ( 0 ) = 0 f(0)=0 and f ( π ) = π f(\pi)=\pi , then 0 π f 1 ( x ) d x = π 2 0 π ( x + sin x ) d x = π 2 2 2 2.9348 \int_0^\pi f^{-1}(x)\,dx=\pi^2-\int_0^\pi (x+\sin x)\,dx=\frac{\pi^2}{2}-2\approx2.9348

Note that, this technique won the 2013 MIT Integration Bee .

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.

aditya vikram - 6 years, 6 months ago

Doing it by the help of graph was easier

Righved K - 5 years, 6 months ago
Ayush Verma
Nov 20, 2014

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 = [ t 2 2 + t sin t + cos t ] 0 π = 2.9348 let\quad x=t+\sin { t } ,dx=\left( 1+\cos { t } \right) dt\\ \\ \Rightarrow f^{ -1 }\left( x \right) =f^{ -1 }\left( t+\sin { t } \right) =t\\ \\ \& x=0\quad \Rightarrow t=0\\ \\ x=\pi \quad \Rightarrow t=\pi \\ \\ \int _{ 0 }^{ \pi }{ f^{ -1 }\left( x \right) dx= } \int _{ 0 }^{ \pi }{ t\left( 1+\cos { t } \right) dt } \\ \\ \quad \quad \quad ={ \left[ \cfrac { { t }^{ 2 } }{ 2 } +t\sin { t } +\cos { t } \right] }_{ 0 }^{ \pi }\\ \\ \quad \quad \quad =2.9348

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.

Keshav Tiwari - 6 years, 6 months ago

Given that f ( x ) = y = x + sin x f(x) = y = x + \sin x and to find I = 0 π f 1 ( x ) d x I = \displaystyle \int_0^\pi f^{-1}(x) \ dx . Swapping x x with y y , then we have x = y + sin y x = y + \sin y and to find I = a b y d x I = \displaystyle \int_a^b y \ dx . And from x = y + sin y x = y + \sin y 0 = a + sin a \implies 0 = a + \sin a a = 0 \implies a = 0 , similarly b = π b=\pi .

x = y + sin y d x = d y + cos y d y = ( 1 + cos y ) d y 0 π x d x = 0 π y d x + 0 π sin y d x 0 π y d x = 0 π x d x 0 π sin y d x = π 2 2 0 π sin y ( 1 + cos y ) d y = π 2 2 + cos y 0 π 0 π sin 2 y 2 d y = π 2 2 2 + cos 2 y 4 0 π = π 2 2 2 + 0 2.935 \begin{aligned} x & = y + \sin y \\ \implies dx & = dy + \cos y \ dy = (1+\cos y) \ dy \\ \implies \int_0^\pi x \ dx & = \int_0^\pi y \ dx + \int_0^\pi \sin y \ dx \\ \int_0^\pi y \ dx & = \int_0^\pi x \ dx - \int_0^\pi \sin y \ dx \\ & = \frac {\pi^2}2 - \int_0^\pi \sin y (1+\cos y) \ dy \\ & = \frac {\pi^2}2 + \cos y \ \bigg|_0^\pi - \int_0^\pi \frac {\sin 2y}2 \ dy \\ & = \frac {\pi^2}2 -2 + \frac {\cos 2y}4 \ \bigg|_0^\pi \\ & = \frac {\pi^2}2 -2 +0 \\ & \approx \boxed{2.935} \end{aligned}

Kyle Finch
Apr 4, 2015

the general property is

a b f ( x ) + f ( a ) f ( b ) f 1 ( x ) = b f ( b ) a f ( a ) \displaystyle \int_{a}^{b} f(x) + \displaystyle \int_{f(a)}^{f(b)} f^{-1}(x)=bf(b)-af(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 ) f^{-1}(x)\ne f(x) here

Pranjal Jain - 6 years, 6 months ago

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