Back to the Calculus

The graphs of $y=f(x)$ and $y=f^{-1}(x)$ are symmetrical in the line $y=x$ :

Using this symmetry, we can find the integral geometrically, as below:

From the diagram, the integral is just $\beta^2-\alpha^2$ . So we need to find natural numbers $\alpha$ and $\beta$ satisfying $\beta^2-\alpha^2=13$ . Factorising, we have $(\beta-\alpha)(\beta+\alpha)=13$ ; since $13$ is prime, we can only factorise it in naturals as $1\times13$ , so the only possible solution is $\alpha=6$ and $\beta=7$ , giving the answer $\boxed{42}$ .

It is given that the function is strictly increasing and it intersects with its inverse at two points. The points of intersection must be located on the line $y = x$ since that is the line of symmetry for the function and its inverse.

This implies: $f(\alpha) = f^{-1}(\alpha) = \alpha$ and $f(\beta) = f^{-1}(\beta) = \beta$

Consider the integral $I = \int_{\alpha}^{\beta}f^{-1}(x) dx$

Taking $f^{-1}(x) =z$ leads to: $x = f(z)$ $dx = f'(z) dz$

When $x = \alpha$ then $z = \alpha$ and the same argument holds for $\beta$ .

The integral becomes:

$I = \int_{\alpha}^{\beta}zf'(z)dz = \int_{\alpha}^{\beta}xf'(x)dx$

Integrating by parts gives:

$I = xf(x)\biggr\rvert_\alpha^\beta - \int_{\alpha}^{\beta}f(x)dx$

Which implies

$\int_{\alpha}^{\beta}f^{-1}(x) dx = \beta f(\beta) - \alpha f(\alpha) - \int_{\alpha}^{\beta}f(x)dx$

Which implies:

$\int_{\alpha}^{\beta}\left(f^{-1}(x) + f(x)\right)dx = \beta f(\beta) - \alpha f(\alpha)$

Which implies:

$\beta f(\beta) - \alpha f(\alpha) = 13$

Using the above argument that $f(\alpha) = f^{-1}(\alpha) = \alpha$ and $f(\beta) = f^{-1}(\beta) = \beta$ , we get:

$\beta^2 - \alpha^2 = 13$

Now, it is given that $\alpha$ and $\beta$ are natural numbers. The only combinations of natural numbers that satisfies that above equation is when the absolute values of $\alpha$ and $\beta$ are 6 and 7 respectively. Hence the answer is 42