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Find the area bounded by y=x+sinx and its inverse between x=0 and x=2π.
A) 2
B) 4
C) 6
D) 8

Answer
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Hint:
Here, we will use the concept of integration to find the area bounded by the curves. First we will find the point of inflection and then integrate the curves to find the area. Integration is the process of adding small areas to find the whole area.
Formula Used: Area between two curves which intersect each other at [a,b] is given by ac[f(x)g(x)]dx+cb[g(x)f(x)]dx

Complete step by step solution:
We are given with a function y=x+sinx.
Replacing the function ywith f(x) , we get
f(x)=x+sinx
Here, x[0,2π].
Differentiating with respect to x, we get
dydx=1+cosx0, xR
Again differentiating with respect to x, we get
d2ydx2=sinx=0
We know that sinx=0 at all integers of π.
x=nπ,nZ
Therefore, x=nπ is the point of inflection for f(x).
So, we have x(0,π), sinx>0 and x+sinx>x.
Also x(π,2π), sinx<0and x+sinx<x.
Using these we can draw a graph as shown below:
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Area between two curves which intersect each other at [a,b] is given by ac[f(x)g(x)]dx+cb[g(x)f(x)]dx.
Now,
 Area bounded by y=x+sinx and its inverse =02π[f(x)f1(x)]dx
Inverse function is symmetrical with respect to y=x.
Thus by using the property of symmetry, we get
Area bounded by y=x+sinx and its inverse =40π[f(x)x]dx
Substituting the function f(x)=x+sinx, we get
Area bounded by y=x+sinx and its inverse =40π[(x+sinx)x]dx
Subtracting the terms in the integrand, we get
Area bounded by y=x+sinx and its inverse =40πsinxdx
Integrating the function, we get
Area bounded by y=x+sinx and its inverse =4[cosx]0π
Substituting the upper limit and lower limit, we get
Area bounded by y=x+sinx and its inverse =4[cosπ(cos0)]
Adding and multiplying, we get
Area bounded by y=x+sinx and its inverse =4[(1)(1)]
Area bounded by y=x+sinx and its inverse =4[1+1]
Adding the terms, we get
Area bounded by y=x+sinx and its inverse =4(2)=8 sq. units

Therefore, the area bounded by y=x+sinx and its inverse between x=0 and x=2π is 8 sq. units.

Note:
We need to keep in mind that, while doing integration, we should notice from the graph the curve at the bottom should be subtracted from the curve at the top. We should also know that the inflection point is a point where a function changes its sign and its direction. We should remember that the function has to be differentiated twice to find the inflection points. The limits of the integral become when both the curves are symmetrical to each other.
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