# Using an integration technique:$\int^b_a \left[f(x)+f^{-1}(x)\right]dx=b^2-a^2$

If $a,b$ are fixed points of $f$, then $$\int^b_a \left[f(x)+f^{-1}(x)\right]dx=b^2-a^2$$

Apparently it was used in the final of the 2013 MIT Integration bee but I can't find that question anywhere..

REPHRASING QUESTION: Could someone show me some integrals that can be cleverly solved with this technique?

• By fixed point, you mean to say that $f(a)=a$? – Simply Beautiful Art Jan 3 '17 at 20:51
• I think he is asking for applications, not a proof .. – Blencer Jan 3 '17 at 20:58
• @Dr.MV you are correct! – bigfocalchord Jan 3 '17 at 20:58
• @Dr.MV Sorry for the noise and thanks for the reminder. – Did Jan 3 '17 at 21:03
• @Did No need to apologize. You're one of the most respectable mathematicians on this site. And Happy New Year Didier! -Mark – Mark Viola Jan 3 '17 at 21:07

If we put $x=f(y)$ with $dx=f'(y)dy$ and integrate by parts, we get

$$\int_a^b f^{-1}(x)dx=\int_a^b yf'(y)dy$$ $$=\big[yf(y)\big]_a^b-\int_a^b f(y)dy$$

$$=bf(b)-af(a)-\int_a^b f(x)dx.$$

$$=b^2-a^2-\int_a^b f(x)dx$$ qed. for example, we can take

$f(x)=\tan(\frac{\pi x}{4}), a=0, b=\pm 1$. or

$f(x)=\sqrt{x^3+1}-1,\; a=0,b=2$

For instance:

$$\int_0^1 \frac{x(x-5)}{(x+1)(x-2)}dx = \int_0^1\left(\frac{2x}{x+1} +\frac{x}{2-x}\right)dx = 1 - 0 = 1$$

Problem: Evaluate $$\int\limits_{0}^{1}{\arcsin x\,dx}.$$ Solution: Let $f(x) = \sin\left(\frac{\pi}{2}x\right)$. Then $0$ and $1$ are fixed points of $f$, and $f$ is increasing (and hence invertible) on $(0,1)$. Since $f^{-1}(x) = \frac{2}{\pi}\arcsin x$, we have $$\int\limits_{0}^{1}{\sin\left(\frac{\pi}{2}x\right) + \frac{2}{\pi}\arcsin x\,dx} = 1.$$ Since $$\int\limits_{0}^{1}{\sin\left(\frac{\pi}{2}x\right)\,dx} = \frac{2}{\pi}\int\limits_{0}^{\pi/2}{\sin u\,du} = \frac{2}{\pi}$$ it follows that $$\frac{2}{\pi}\int\limits_{0}^{1}{\arcsin x\,dx} = 1-\frac{2}{\pi} \implies \int\limits_{0}^{1}{\arcsin x\,dx} = \boxed{\frac{\pi}{2}-1}.$$

Of course, $\arcsin x$ does have an elementary antiderivative which can be found using integration by parts, but this method seems faster to me.

There are a few things going on here: $f$ is implicitly assumed to be continuous, and since $f$ is invertible it must be monotone. You may as well assume that $a < b$. Then $f(a) = a < b = f(b)$ so that $f$ is increasing.

If $f$ is differentiable you can make the substitution $s = f^{-1}(t)$, $f'(s) \, ds = dt$ to find $$\int_a^b f^{-1}(t) \, dt = \int_a^b s f'(s) \, ds$$ and then you can integrate by parts to find $$\int_a^b s f'(s) \, ds = b^2 - a^2 - \int_a^b f(s) \, ds.$$

• It also has a nice graphical interpretation. – Simply Beautiful Art Jan 3 '17 at 20:56
• I think he is asking for applications, not a proof .. – Blencer Jan 3 '17 at 20:58