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Why $\int _c^df^{-1}\left(y\right)\:dy+\int _a^b\:f\left(x\right)dx=b\cdot d-a\cdot c$ ? where f is an bijective function and $f(a)=b,f(c)=d,$ I don't understand graph... I can't see on graph this equality, so have somebody patience to explain me on graph this equality?

I think it is not duplicate, because he wants a rigorous proof, not a draw, and I want to understand from graph, I think is not duplicate... I don't know why they think is duplicate

Yiorgos S. Smyrlis
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Lucas
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    If $f(x)=x$ the left hand side is $$\dfrac{1}{2}((d^2+b^2)-(c^2+a^2))$$ which is clearly not the same as the right-hand side. For instance, take $d=1$, the rest 0. Is there some additional assumption you are not relaying? – David P Apr 20 '15 at 21:58
  • can you explain on graph, this equality? – Lucas Apr 20 '15 at 22:01
  • @Lucas: You forgot to stipulate $f(a) = b, f(c) = d$, as is stated in the Wikipedia article. – Brian Tung Apr 20 '15 at 22:04
  • Maybe edit your post then to include the assumptions $f(a)=c$, $f(b)=d$, and that $f$ is continuous. – David P Apr 20 '15 at 22:04
  • @pizza this is not duplicate, because he want a rigorous proof, not a draw, and I want to understand from graph, I think is not duplicate... I don't know why they think is duplicate – Lucas Apr 21 '15 at 05:00
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    There is a graph here –  Apr 21 '15 at 05:06

2 Answers2

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Without reference to a "graph," note that if we have $f(a)=c$ and $f(b)=d$, then

$$\begin{align} \int_a^b f(x) \,\,dx+\int_c^d f^{-1}(y) \,\,dy &=\int_a^b f(x) \,\,dx+\int_a^b f^{-1}(f(x)) f'(x) \,\,dx\\\\ &=\int_a^b f(x) \,\,dx+\int_a^b x f'(x) \,\,dx\\\\ &=\int_a^b f(x)+x f'(x) \,\,dx\\\\ &=\int_a^b (xf(x))' \,\,dx\\\\ &=bf(b)-af(a)\\\\ &=bd-ac \end{align}$$

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Mark Viola
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  • please... can you make me to understand graph? I suppose $\int _c^d:f\left(y\right)dy:+:\int _a^b:f\left(x\right)dx$ is all who's mark with red – Lucas Apr 20 '15 at 22:08
  • Sure. For the graph, the shaded areas are give by $b \times d$ less the small "box", which has area $a \times c$. Does that help? – Mark Viola Apr 20 '15 at 22:11
  • The answer is $bd - ac$. Since $b>a$ and $d>c$, then $bd - ac>0$. – Mark Viola Apr 20 '15 at 22:17
  • so $b\cdot d$ is all areas, excepting $a\cdot c$ ? – Lucas Apr 20 '15 at 22:20
  • $b \times d$ is the area of the "big box" and $a \times c$ is the area of the "small box." The difference is the sum of the areas under (1) $f$ from $a$ to $b$ and (2) $f^{-1}$ from $c$ to $d$. – Mark Viola Apr 20 '15 at 22:22
  • after $b\cdot d:-:a\cdot c$ how will see graph? all red? the "small box" will be red? – Lucas Apr 20 '15 at 22:24
  • The "graph" of the sum of the areas will look like a "big box" with a smaller box cut from its lower corner. – Mark Viola Apr 20 '15 at 22:26
  • and the red stripe between f and f^-1 will disappear? – Lucas Apr 20 '15 at 22:29
  • That "stripe" represents $f$ and the image of $f^{-1}$. – Mark Viola Apr 20 '15 at 22:31
  • the "small box" includes white area ? – Lucas Apr 20 '15 at 22:33
  • Yes, that is correct. – Mark Viola Apr 20 '15 at 22:34
  • why $b\cdot d$ is not a $Point:\left(b,d\right)$ ? – Lucas Apr 20 '15 at 22:36
  • One multiplies $b$ and $d$ to find an area. So, $b \times d$ is just a number. The physical interpretation of that one number is the amount of space a box of length $b$ and height $d$ occupies. Make sense now? – Mark Viola Apr 20 '15 at 22:43
  • yes, I was confused with xOy axis when denote coordinates, but I suppose we can note $b\cdot d:=:P\left(b,:d\right)$ – Lucas Apr 20 '15 at 22:46
  • however thank you with all my heart, you are a patient and good person and I am wondering because there exist people like you... nice to meet you and good luck in what you do ! you was my hero, because tomorrow I'll take an examination – Lucas Apr 20 '15 at 22:52
  • You are more than welcome! Your kind words mean a lot to me ... more than you know. It has been my pleasure - just glad I was able to help. – Mark Viola Apr 20 '15 at 22:54
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Not true!

Take $f(x)=x$ - it is a bijection.

Then $f^{-1}(x)=x$, and $$ \int_c^d f^{-1}(y)\,dy+\int_a^bf(x)\,dx=\frac{1}{2}(d^2-c^2)+\frac{1}{2}(b^2-a^2) \ne bd-ac. $$ However, if $c=f(a)$, and $d=f(b)$, then it holds! Simply draw the figure of the graph of $f$. Then $f^{-1}$ is also visible in the same graph (reflection along $y=x$). Then $bd-ac$ is the area under $f$ plus the area under $f^{-1}$ (reflected) in the interval $[a,b]$ in the $x$ direction.

Yiorgos S. Smyrlis
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