# Integrals and bijective functions

Assume that $$f: [a, b] \rightarrow [a,b]$$ is continuously differentiable bijection and that $$f(a) = a$$, $$f(b) = b$$.

Show that

$$\int_a^b g(x) \, dx + \int_a^b f(x) \, dx = b^2-a^2$$

where $$g:[a, b] \rightarrow [a, b]$$ is the inverse function of $$f$$

With this one I have no idea where to start. What should we do with the information about $$f$$ being bijective and $$g$$ being the inverse of $$f$$?

• Do you know that you wrote {\int_{{a}}^{{b}}}{{g(x)}}{d}{x}}}} where it would suffice to write \int_a^b g(x)\,dx? $\qquad$ – Michael Hardy Feb 16 at 17:33
• Your result holds under the more general condition that $f$ is strictly increasing on $[a, b]$. – Paramanand Singh Feb 17 at 2:47
• Does this answer your question? integration of sum of $f$ and its inverse – Martin R Feb 17 at 9:09

Let $$u=f^{-1}(x)$$. Then $$x=f(u)$$ and $$dx=f'(u)\,du$$. We will integrate by parts:

$$\int_a^b g(u) h'(u)du = \left[g(u)h(u)\right]_a^b-\int_a^b g'(u)h(u)du$$

by choosing $$g(u)=u$$ and $$h(u)=f(u)$$, so:

\begin{aligned} \int_a^b u f'(u)du &= \left[uf(u)\right]_a^b-\int_a^b 1\cdot f(u)du\\ &=bf(b)-af(a)-\int_a^b f(u)\,du \end{aligned}

Since $$f(b)=b$$ and $$f(a)=a$$, we have:

$$bf(b)-af(a)=b^2-a^2$$

Chaining all this together:

\begin{aligned} \int_a^b f^{-1}(x)\,dx&=\int_a^b uf'(u)\,du\\ &=\int_a^b u(f(u))'\,du\\ &= \left[uf(u)\right]_a^b-\int_a^b f(u)\,du\\ &=b^2-a^2-\int_a^bf(x)\,dx \end{aligned}

from which we deduce the conclusion.

• There's something Im not seeing here. Im used to using this $\int udv = uv - \int v du$ for integrating by parts. What is our $dv$ here? – Daniel Feb 16 at 16:32
• @Daniel, that's $f'(u)$. In your notation $(u,v)$ correspond to $(u,f(u))$ in mine. – LHF Feb 16 at 16:37
• I changed $f'(u)du$ and $f(x)dx$ to $f'(u)\,du$ and $f(x)\,dx.$ The latter is standard. – Michael Hardy Feb 16 at 17:34
• @Atticus Is there any chance you could include a couple more steps to this? Im not sure I see where you applied integration by parts and how did $\left[uf(u)\right]_a^b = b^2 - a^2$. – Daniel Feb 17 at 17:06
• @Daniel, I added some detail. I hope it's more clear now. – LHF Feb 17 at 17:16

Suppose for convenience $$0 Draw a picture. The area between the graph of $$f$$ and the $$x$$-axis is $$\int_a^b f(x)\, dx.$$ Same for $$g$$ with respect to the $$y$$-axis. If we add these two areas, we get the area of $$[0,b]\times [0,b]$$ minus the area of $$[0,a]\times [0,a],$$ which is $$b^2-a^2.$$

Here's a graphical hint (not a proof of course): • That's exactly what I was doing. Nice picture though. – zhw. Feb 17 at 16:07