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$f:[0,\infty) \rightarrow [0,\infty)$ is strictly increasing. $I(a) = \int_0^a f(x) dx$ and $J(b) = \int_0^b f^{-1}(x) dx$.

Which must be correct?

1) $J(b)$ is equal to the area bounded by $x=0,y=b,f(x)$.

2) $a>0$ and $0<b<f(a)$ implies $I(a)+J(b)<ab$.

3) $a>0$ and $b>f(a)$ implies $I(a)+J(b)>ab$.

Attempt

Since the inverse function is the reflection of $f(x)$ in the line $y=x$, point 1) should be correct? Then, point 2) and 3) follows from point 1)?

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Does 1) actually define a boundary? The right side seems open. Why are the $>0$ conditions necessariy, when $f$ isn't even defined for $a,b<0$? –  NikolajK Oct 29 '12 at 9:14
    
(1) is true if it actually reads $x=0,y=b,y=f(x)$; as it stands, no region is defined. And yes, (2) and (3) are true. –  Brian M. Scott Oct 29 '12 at 9:16
1  
@BrianM.Scott: Graphically, I'd say 2) is false. –  NikolajK Oct 29 '12 at 9:21
    
@Nick: Sorry, you’re right: I was thinking of $af(a)$, not $ab$. –  Brian M. Scott Oct 29 '12 at 9:28
    
@BrianM.Scott: Apology accepted. –  NikolajK Oct 29 '12 at 9:31

1 Answer 1

up vote 1 down vote accepted

The area in 1) is not bounded by the three relations. So it cannot be calculated, and is in any case not equal to $J(b)$.

Edit: Now that the conditions in (1) has been corrected, it is possible to determine whether or not it is true. And it is, given some assumption like the one in my comment below, namely that if $f(0) = c > 0$, then we set $f^{-1}(y) = 0$ for all non-negative $y<c$. End edit

If $b = f(a)$, then $I(a)$ and $J(b)$ will be the area of two parts of the rectangle with corners $(0, 0), (a, 0), (a, b), (b, 0)$, so in that case $I(a) + J(b) = ab$.

(2) is false, and (3) is true, since if $b\neq f(a)$, then $ab$ will be smaller than the areas of the two integrals. You really should draw a picture of the whole setting, just to convince yourself of this.

A written reasoning for the above paragraph can be done as follows: Assume $b<f(a)$, as is the hypothesis for (2). (3) follow very similar reasoning. Then the rectangle with sides parallel to the axes and with diagonal from the origin to $(a, b)$ will be divided into two by the graph of $f$. The area above the graph is $J(b)$, but the area below will be smaller than $I(a)$, since the contribution to $\int_0^af(x)dx$ from all $x$-values larger than $f^{-1}(b)$ will be limited to height $b$, which is smaller than the height of the graph. So we have $I(a) + J(b) = ab + \int_{f^{-1}(b)}^a(f(x) - b)dx > ab$

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Thanks for the help. Question: Is 1) a theorem? (with the corrected $y=b$ condition) Are there conditions like non-continuity, negativity etc that will violate this? Does this work for all connected real intervals $(a,b)$? For some reason I can't wrap my head around this "fact". –  Legendre Oct 29 '12 at 13:45
    
With the corrected conditions in (1), it is true. Also, strictly increasing functions are continuous almost everywhere, and thus integrable. You could say that (1) formulates the definition of the indefinite integral $\int f^{-1}(b)db$, give or take a constant term. The only care you have to take in that situation is that if $f(0) = c > 0$, then you need to say something along the lines of "$f^{-1}(y) = 0$ for all non-negative $y<c$". –  Arthur Oct 29 '12 at 14:23
    
Thank you so much. :) –  Legendre Oct 30 '12 at 6:05

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