0

I am confused this problem is really hard for me . I don't know how to calculate this function's inverse but my teacher said that this can be done without finding out the inverse of the function . Please help .

Elaqqad
  • 13,725
user169772
  • 71
  • FYI the function looks like this: http://www.wolframalpha.com/input/?i=sketch+y%3Dx%2Bsinx%2C+y%3Dx, and its inverse is just reflection of the graph in the line $y=x$. Now refer to answer below – user160738 Mar 21 '15 at 13:49
  • There are infinitely many regions between those two functions, and half the time one function is above and the other half the other function is. Do you want the area of just one of those regions? If so, Elaqqad's answer suffices and is a good one. If you want the signed area of two of those regions, the answer is zero. – Rory Daulton Mar 21 '15 at 13:49

2 Answers2

3

Hint The function and its inverse will be symmetric about the line $y=x$. so you can mange your calculation to computing a simple integral.

For example if the bounds are $x=0$ and $x=\pi$ then the wanted area is just double the area between $y= x+\sin x$ and $y=x$

Community
  • 1
Elaqqad
  • 13,725
0

First of all, it's reasonable to take into account that the graphs of the function and of its inverse are symmetric with respect to the $y=x$. If it's nessecary to find the exact area, for example from $0$ to $\pi$, the answer is $\int_{0}^{\pi}{((x+\sin(x))-(x)) \cdot dx}=\int_{0}^{\pi}{\sin(x)}=-\cos(x) | _{0}^{\pi}=-((-1)-(1))=2$

hyperkahler
  • 2,921