How can I find the area between the graphs of a function and its inverse? I have the following function $$f(x)=x\cdot e^{x^2-1} $$ and I want to find the are between this function and its inverse. I'm not sure how to calculate the integral because I know that for this type of problem I need to find where the two functions intersect.
 A: This is an odd function, so its graph to the left of $x=0$ is a $180^\circ$ rotation of its graph to the right of $0$. On the interval $(0,\infty)$ the functions $f,f',f''$ are everywhere positive. So the graph of $f$ cannot intersect the graph of $y=x$ anywhere except at $(0,0)$ and $\pm(1,1).$
Since $f$ is concave upward on $[0,1],$ the graph of $f^{-1}$ is concave downward on that interval. Since they are reflections of each other about the line $y=x$, the area above $y=f(x)$ and below $y=x$ is the same as the area above $y=x$ and below $y=f^{-1}(x).$
Therefore
$$
\Big(\text{area between $f$ and $f^{-1}$ to the right of $0$}\Big) = 2\int_0^1 (x - f(x))\,dx.
$$
And
$$
\int_0^1 f(x)\,dx = \frac 1 2 \int_0^1 e^{x^2-1} \Big( 2x\,dx\Big) = \frac 1 2 \int_{-1}^0 e^u \,du.
$$
A: HINT: The functions intersect where $f(x)=x$.
HINT2: This is at $-1, 0, 1$
HINT3: The inverse function is the function reflected around the line $y=x$, so you find the intersections of $f(x)$ and $f^{-1}(x)$ when $f(x)$ intersects with $y=x$. You use this to integrate and find the area between $f(x)$ and $x$, then multiply by $2$ for symmetry.
