A Challenging Problem for You!

Calculus Level 2

A r e a b o u n d e d b y c u r v e f ( x ) = x + s i n x a n d i t s i n v e r s e f u n c t i o n b e t w e e n o r d i n a t e s x = 0 t o x = 2 π i s ? Area\quad bounded\quad by\quad curve\quad f(x)=x+\quad sin\quad x\quad and\quad its\quad inverse\quad function\quad between\quad ordinates\quad x=0\quad to\quad x=2\pi \quad is? Give a solution if you can. Follow for more


The answer is 8.

Hrisheek Yadav by Hrisheek Yadav , 20, India

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1 solution

Otto Bretscher
Oct 28, 2015

By symmetry, this is twice the area beween sin ⁡ x + x \sin{x}+x and x x for 0 < x < 2 ∗ π 0<x<2*\pi or four times the area between sin ⁡ x + x \sin{x}+x and x x for 0 < x < π 0<x<\pi . Thus the area is 4 ∫ 0 π sin ⁡ x d x = 8. 4\int_0^{\pi}\sin{x}dx=\boxed{8.}

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