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# Sir, Pls solve this question.

Asked by rsudipto 8th December 2018, 2:55 PM
Plotting
put x = 0 in x + sinx, you get 0
put x = pi, you get pi, which is  3.14
then x = 2pi, you get 2pi, which is 6.28
make a rough diagram

now draw a graph mirror to the blue one, which will be the inverse (red)
inverse graph can be made by taking mirror about y = x

Method 1
now this inverse graph is x - sinx graph.......this we can check by sub the values like pi , 2pi etc (OR refer to method 2)

The function f(x) = x+sin(x) and its inverse f '(x) = x-sin(x) are plotted in figure.
It can be seen that inverse of f(x) is reflection about the line y=x

Let us calculate area bounded between f(x) and iis inverse between x=0 and x=π .

Hence area bounded between f(x) and iis inverse between x=0 and x= 2π is  8 sq. units

Method 2

we can find the area enclosed by x + sinx and x between 0 to pi (shaded part) and multiply it with 4

Answered by Expert 9th December 2018, 3:43 AM
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