Area inside a region

An taking the inverse of a function is same as reflection of its graph along the y=x axis. So the area would be the twice the value of the area between the curve and the line y=x $A = 2 \times \int_{0}^{\pi} x + sin(x) dx - \pi^{2} = 4$

The first thing to notice is that that
$y=f^{-1}(x) \Leftrightarrow x = f(y)$

So, let $A$ be the set of points $(x_0,y_0) \in \{(x,y):y=f(x)\}$ and $B$ be the set of points $(x_0,y_0) \in \{(x,y):y=f^{-1}(x)\}$ .

With that in mind, it is possible to conclude that $(x_0,y_0) \in B \Leftrightarrow (x_0,y_0) \in \{(x,y):x=f(y)\}$

$(x_0,y_0) \in B\Leftrightarrow (y_0,x_0) \in A$

If we want to find the points in which $A$ and $B$ that intersect with each other, we know that $(x_0,y_0) \in B, (x_0,y_0) \in A$ , which implies $(y_0,x_0) \in A, (x_0,y_0) \in A$ . So, $x_0 = y_0$ .

So if $y= f(x) = f^{-1}(x)$ , then $y=x$

In this case, $x+sin(x) = x$

$sin(x) = 0$

$x = k\pi$ , k is an integer.

If we want one region, let us just take two adjacent points belonging in $A$ and $B$ , say $x=0$ and $x=\pi$ . We know that since both curves are continuous, there must be a closed region in the interval between $0$ and $\pi$ .

It is important to notice that the two are a reflection of each other with respect to the line $y=x$ .

That makes it easier for us to calculate, since it is clear that the area between $y=x$ and $y = f(x) = x + sin(x)$ is always half the area of the region.

$\frac{A}{2} = \|\int_{0}^{\pi} (f(x)-x) dx\|\ \\ = \int_{0}^{\pi} ((sin(x)+x)-x) dx \\ = \int_{0}^{\pi} sin(x) dx \\ = 2 \\ \Rightarrow A ={2}\times{2} = 4$

If $f(x)=f^{-1}(x)$ , then immediately follows that $f(x)=x \wedge f^{-1}(x)=x$ (for the values of $x$ at intersections of those two functions), which is obvious if you think about it: the inverse function is the function of x mirrored in the line $y=x$ and intersections between the function and the inverse of that same function will be on the line $y=x$ because on those points the mirrored point will be at the same spot. So we can immediately conclude that $f(x)=x$ and thus $sin(x)=0$ which implies that $x=k \cdot \pi \: | \:k \in \mathbb{Z}$

Now we know for which $x$ there will be a intersection, we will just need two following values for $x$ . Let's take $x=0$ and $x=\pi$ , for example. We can use that the area within the function and the inverse of that function for $0 \leq x \leq \pi$ will be $2 \cdot |\int_{0}^{\pi} (f(x)-x) dx|$ . This is because the area between the two functions is double the area of $f(x)$ to the line $y=x$ , because the inverse function still is mirrored in the line $y=x$ .

This means that the asked area has to be $2 \cdot |\int_{0}^{\pi} (sin(x)+x-x) dx| = 2 \cdot |\int_{0}^{\pi} (sin(x)) dx| = 2 \cdot |[-cos(x)]_0^{\pi}|= 2 \cdot|1-(-1)| = 2 \cdot 2 = 4$

Consider the following graph:

Where we present (red) $f(x)$ and (blue) $f^{-1}(x)$ trapped inside a square of side $\pi$ . (The two intersections happen at $(0,0), (\pi, \pi)$ ). It is clear that $\int_0^\pi x + \sin x dx = B + C$ so $B + C = \frac{\pi^2}{2} - 2$ . But we know that $A + B + C = \pi^2$ and because of a symmetry argument already presented in another solution, $C = A$ so we have the system $2A + B = \pi^2 \\ B + A = \frac{\pi^2}{2} - 2$ It is solvable and we really just care for $B = 4$ .