If f(x)=x+sinx , then the value of ∫π2πf−1(x)dx is

I=∫π2πf−1(x)dxPutting x=f(t) or dx=f'(t)dt , we get I=∫π2πt.f'dt =2πf(2π)−πf(π)−|t22−cost|π2π=3π22+2

If f(x)=x+sinx , then the value of ∫π2πf−1(x)dx is

I=∫π2πf−1(x)dxPutting x=f(t) or dx=f'(t)dt , we get I=∫π2πt.f'dt =2πf(2π)−πf(π)−|t22−cost|π2π=3π22+2

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If $f (x) = x + \sin x$ , then the value of $\int_{π}^{2 π} f^{- 1} (x) d x$ is

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$\frac{3 π^{2}}{2} - 2$

$\frac{π^{2}}{2} - 2$

$\frac{3 π^{2}}{2} + 2$

$\frac{π^{2}}{2} + 2$

answer is C.

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$I = \int_{π}^{2 π} f^{- 1} (x) d x$
Putting $x = f (t)$ or $d x = f' (t) d t$ , we get $I = \int_{π}^{2 π} t . f' d t$
$= 2 π f (2 π) - π f (π) - {| \frac{t^{2}}{2} - \cos t |}_{π}^{2 π} = \frac{3 π^{2}}{2} + 2$

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If f(x)=x+sinx , then the value of ∫π2πf−1(x)dx is

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If $f (x) = x + \sin x$ , then the value of $\int_{π}^{2 π} f^{- 1} (x) d x$ is