definite integration

f(x)=sin x + x ; find ∫0pi f-1(x) dx.

12 Answers

1
xYz ·

nice one !
answer is pi^2/2 -2 ??

1
ut10 ·

how ,please post solution

106
Asish Mahapatra ·

sorry if bit too unclear

RED line is f-1(x) . Black line is f(x)

1
ut10 ·

thank you friends for the solution, but i did it in another way. tell me if my solution is correct
i derived f-1(x)=x-sinx and then integrated it over 0 to pi.

1
xYz ·

106
Asish Mahapatra ·

f-1(x) is NOT x-sinx

although it seems to be...
but it is not

1
ut10 ·

but if we draw g(x) =x-sinx it comes out to be same as f-1(x).

106
Asish Mahapatra ·

no it doesnt there are a few subtle differences..

This was discussed sometime back and answered quite nicely by kaymant sir... Let me search for it

39
Pritish Chakraborty ·

We can't find f-1(x) that way.

We know that fof-1(x) = x
That means sin(f-1(x)) + f-1(x) = x
I don't think you can isolate f-1(x) from here.

1
ut10 ·

thanks, i found graph of f(x)=x+sin(x) (blue) and its inverse(red)

1
ut10 ·

and graph of g(x)=x-sin(x) (blue) and its inverse(red)

341
Hari Shankar ·

It is easy to see that f^{-1} (0) = 0; f^{-1} {\pi} = \pi

We have to find \int_0^{\pi} f^{-1} (x) \ dx

Let x = f(y).

Then the integral is \int_0^{\pi} y (1+\cos y) \ dy.

By partial integration you get \frac{\pi^2}{2} - 2

There's some theorem with integrals of f(x) and f-1(x) which was discussed in this forum long back which I suspect simplifies these problems. I'll try to locate that thread

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