Area between two inverse function? Consider the functions $f, g:[0,1] \rightarrow[0,1]$ given by
$$
f(x)=\frac{1}{2} x(x+1) \text { and } g(x)=\frac{1}{2} x^{2}(x+1)
$$
Then the area enclosed between the graphs of $f^{-1}$ and $g^{-1}$ is
My approach
inverse function means it is the
reflection along the line y=x.
Reflection does not change the area under the curve
$$
\text { So area between two curve }  \int_{0}^{1}[f(x)-g(x)]=\int_{0}^{1}\left[f^{-1}(x)-g^{-1}(x)\right] \\
 =\frac{1}{2}(1 / 2-1 / 4)=\frac{1}{8}
$$

*ps:- If i do further calculation ,more precisely then   the quadratic eq vertex point will be $(-b/2a,-d/4a)$ i.e $(-1/2,-1/8)$.....
Please correct me,if I am wrong
 A: You are correct.  Another way to see that this is true is to look at the graphical approach to integrating the difference in area bounded by the inverses of $f(x)$ and $g(x)$.

$\int_0^1 g^{-1}(x)$ is the area between the y-axis and $g(x)$.  This is equal to the difference in area of the rectangle with coordinates $(0,0), (0,1), (1,0),$ and $(1,1)$ and the integral of $\int_0^1 g(x) \; dx$, or $1-\frac{7}{24}=\frac{17}{24}$.
Repeat this for $f(x)$ to get that $\int_0^1 f^{-1}(x) \; dx=\frac{7}{12}$.  Thus, the area between $f^{-1}(x)$ and $g^{-1}(x)$ on $x=[0,1]$ is $\frac{17}{24}-\frac{7}{12}=\frac{1}{8}$.
$$ $$
This can be applied to other problems with determining the area bounded between $f^{-1}(x)$ and $g^{-1}(x)$.  For the special case where $f(x)$ and $g(x)$ intersect at the origin:
$$\int_a^b f^{-1}(x)-g^{-1}(x) \; dx=\int_a^b \left(f(b)b-f(a)a-f(x)\right)-\left(g(b)b-g(a)a-g(x)\right) \; dx$$
Simplifying this expression with $f(a)a=g(a)a=0$ and $f(b)b=g(b)b$:
$$\int_a^b f^{-1}(x)-g^{-1}(x) \; dx=\int_a^b g(x)-f(x) \; dx$$
However, note that area is positive so the area bounded between $f(x)$ and $g(x)$ and $f^{-1}(x)$ and $g^{-1}(x)$ is equal.
