Show: $ f(a) = a,\ f(b) = b \implies \int_a^b \left[ f(x) + f^{-1}(x) \right] \, \mathrm{d}x = b^2 - a^2 $

Question

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

In the words of 2014 MIT Integration Bee Champion (Carl Lian), the above property was responsible for the champion's victory in the 2013 MIT Integration Bee. How does one go about proving this property?

Related: http://math.stackexchange.com/questions/106509/is-int-ab-fx-dx-int-fafb-f-1x-dy — JimmyK4542, Aug 23 '14 at 01:10
Do you know that $f(a) = a$ and $f(b) = b$ for this problem? — JimmyK4542, Aug 23 '14 at 01:12
Is there no condition on $f$? It seems unlikely to be true for any $f$. — Thomas Andrews, Aug 23 '14 at 01:12
You guys are absolutely right, I missed the fact that $a,b$ must be fixed points of $f$. Thank you! — Ahaan S. Rungta, Aug 23 '14 at 01:13
Also related: http://math.stackexchange.com/questions/562960/prove-that-int-abfxdx-int-fafb-f-1ydy-bfb-afa — JimmyK4542, Aug 23 '14 at 01:14
Uou obviously need $f(x)$ $1-1$ and onto for the question to even make sense, of coruse. — Thomas Andrews, Aug 23 '14 at 01:18
It seems the links @JimmyK4542 posted answer my question. If you want to convert the comments to a full answer, I'll upvote/accept. — Ahaan S. Rungta, Aug 23 '14 at 01:19
Yes, this is essentially a visual result. Curious how it works out if $a<0<b$, visually, but $0\leq a<b$ is visually obvious. — Thomas Andrews, Aug 23 '14 at 01:19

score 13 · Accepted Answer · answered Aug 23 '14 at 01:27

Let's assume that $f$ is monotonic and differentiable. Using the substitution $y = f(x)$ followed by integration by parts, we get:

$\displaystyle\int_{f(a)}^{f(b)}f^{-1}(y)\,dy = \int_{a}^{b}xf'(x)\,dx = \left[xf(x)\right]_{a}^{b} - \int_{a}^{b}f(x)\,dx = [bf(b)-af(a)] - \int_{a}^{b}f(x)\,dx$.

Therefore, $\displaystyle\int_{a}^{b}f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = bf(b)-af(a)$.

If we also know that $f(a) = a$ and $f(b) = b$, this becomes: $\displaystyle\int_{a}^{b}\left[f(x)+f^{-1}(x)\right]\,dx = b^2-a^2$.

score 7 · Answer 2 · edited Jun 02 '23 at 19:45

7

Geometric method

The area of the two colored regions are $b^2-a^2$ (difference of areas of two rectangles)

$$b(b-a)+a(b-a)=b^2-a^2$$

edited Jun 02 '23 at 19:45

Kamal Saleh

6,497

answered Dec 24 '14 at 12:13

Transcend97

87

Kamal Saleh · Answer 3 · 2023-06-02T19:44:57.427

0

It is known that $$\int f^{-1}(x)dx=xf^{-1}(x)-F(f^{-1}(x))+C$$Where $F(x)$ is an antiderivative of $f(x)$. Substituting, we get $$I=\int_a^bf(x)+f^{-1}(x)dx=F(b)-F(a)+bf^{-1}(b)-F(f^{-1}(b))-af^{-1}(a)+F(f^{-1}(a))$$Since $f(a)=a$ and $f(b)=b$, $f^{-1}(a)=a$ and likewise for $b$. So, substituting this in: $$I=F(b)-F(a)+b^2-F(b)-a^2+F(a)=b^2-a^2$$

edited Jun 02 '23 at 19:44

answered Jun 01 '23 at 19:00

Kamal Saleh

6,497

Show: $ f(a) = a,\ f(b) = b \implies \int_a^b \left[ f(x) + f^{-1}(x) \right] \, \mathrm{d}x = b^2 - a^2 $

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