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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

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 $a,b<0$?2012-10-29
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    (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.2012-10-29
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    @BrianM.Scott: Graphically, I'd say 2) is false.2012-10-29
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    @Nick: Sorry, you’re right: I was thinking of $af(a)$, not $ab$.2012-10-29
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    @BrianM.Scott: Apology accepted.2012-10-29
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    !!! Yes, the boundary is supposed to be $y=b$.2012-10-29
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    @NickKidman - Perhaps to avoid $a=0$ case where $ab=0$.2012-10-29

1 Answers 1

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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 $yEnd 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 $b ab$

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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".2012-10-29
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    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 $y2012-10-29