How Does the Integral of an Inverse Function Relate to Areas Under Curves?

[MathematicalPhysicist]

let f be a monotone function on the interval [a,b].
f(a)=\alpha f(b)=\beta.
let g be the inverse function of f i.e f^-1.
prove that:
\int_{\alpha}^{\beta}g(y)dy=b\beta-a\alpha-\int_{a}^{b}f(x)dx when you consider the definition of the integral as area.

here what i did:
Sn=\sum_{i=0}^n g(\zeta_i)(y_i-y_{i-1})
f(x)=y f^-1(y)=x=g(y)
y_i-y_i-1=f(x_i)-f(x_i-1)
S'n=\sum_{i=0}^n \zeta_i(x_i-x_{i-1})
S'_n+\sum_{i=0}^n x_{i-1}\zeta_i=\sum_{i=0}^n x_i\zeta_i=\sum_{i=0}^n g(\zeta_i)y_{i-1}
as n approaches infinity S'_n approaches the second integral in with limits a and b, and the first sum of yi times g(zeta_i) minus zeta_i times x_i-1 converges to b*beta- a*alpha, is this correct?

 

[benorin]

Recall that the inverse of a function is (graphically) the reflection of the original function about the line y=x, and that b\beta and a\alpha are the areas of two rectangles so b\beta-a\alpha is the area of a 'box-shaped' L region. Plot it with say f(x)=x^2.

 

[quasar987]

I think the specification "when you consider the definition of the integral as area" means that you can use the geometric arguments benorin is hinting about. So spoil yourself while you can! :P

 

[0rthodontist]

I'm confused by this. For example why is the last equality true (mainly I am confused as why the subscript on the y is i-1 and not i) and what is "the first sum of yi times g(zeta_i) minus zeta_i times x_i-1 converges to b*beta- a*alpha"? Also I'm going to use just x_i and y[/sub]i[/sub] instead of the zetas. (by the way your limits should probably go from 1 to n, not that it matters much but you'd have a negative subscript on the i-1)

S_n = \sum_{i=1}^n g(y_{i-1})(y_i-y_{i-1}) (evaluating g on the left of each interval)
S_n' = \sum_{i=1}^n f(x_i)(x_i-x_{i-1}) (evaluating f on the right of each interval)
where the x_i are chosen so that g(y_i) = x_i, and x_0 = a, x_n = b, y_0 = alpha, y_n = beta.

I chose to evaluate the function on the left of each interval for g and on the right of each interval for f because I drew a picture which suggested it:
http://img152.imageshack.us/img152/7889/evalqs3.png

S_n = \sum_{i=1}^n x_{i-1}(y_i-y_{i-1})
= \sum_{i=1}^n x_{i-1} y_i - \sum_{i=1}^n x_{i-1} y_{i-1}
S_n' = \sum_{i=1}^n y_i(x_i-x_{i-1})
= \sum_{i=1}^n x_i y_i - \sum_{i=1}^n x_{i-1} y_i

Now, note that b \beta - a \alpha can be written as
\sum_{i=1}^n x_i y_i - x_{i-1} y_{i-1}
because the interior sums cancel and you're left with just x_n y_n - x_0 y_0 = b * beta - a * alpha. Then you can see that Sn + Sn' = b beta - a alpha.

 

[StatusX]

Strictly speaking, you would need to prove that as max(y_i-y_i-1)->0, so does max(x_i-x_i-1), the condition for the Riemann sum to converge to the integral. This can be done in a line using the assumption of uniform continuity, which in turn can be derived from the continuity of the function on a closed interval by the Heine-Cantor theorem.

 

[WigneRacah]

let f be a monotone function on the interval [a,b].
f(a)=\alpha f(b)=\beta.
let g be the inverse function of f i.e f^-1.
prove that:
\int_{\alpha}^{\beta}g(y)dy=b\beta-a\alpha-\int_{a}^{b}f(x)dx when you consider the definition of the integral as area.

Try the substitution:

y=f(x).EDIT: I suppose f is a "regular" function.

 

[MathematicalPhysicist]

0rthodontist, i didnt quite understand what does your illustration convey here?
i used here zetas, because according to the defintion the points which we choose for f(x) should be inside the cell, i.e between x_i-1 and x_i, and not necessarily one of these end points of the divided cell.

 

[0rthodontist]

0rthodontist, i didnt quite understand what does your illustration convey here?

I have shown examples of S3 and S3', with S3 written sideways along the y-axis and S3' written normally. The sum of these can be seen in the diagram as the area b beta - a alpha-the sums interlock so you don't have to go to the limit to get the sum right.
StatusX's point is taken if you want to do it this way. You would have to prove that the norm of the partition for S3' converges to 0.

i used here zetas, because according to the defintion the points which we choose for f(x) should be inside the cell, i.e between x_i-1 and x_i, and not necessarily one of these end points of the divided cell.

You can choose any points you like in the intervals, and it can be one of the end points of the cell. But that doesn't have anything to do with how you denote your choice. zetas would work as well as x's.

 

[MathematicalPhysicist]

i have another question:
prove that if f(x) is continuous and f(x)=\int_{0}^{x}f(t)dt then f(x) is zero.

what i did so far is:
for every e, there exists d, such that whenever |x-x1|<d, |f(x)-f(x1)|<e
|\int_{0}^{x}f(t)dt-\int_{0}^{x_1}f(t)dt|=|\int_{x_1}^{x}f(t)dt|&lt;e.
for every e, there exists d, such that whenever |x-x2|<d |f(x)-f(x2)|<e
we get that |x1-x2|<2d |f(x1)-f(x2)|<2e if x2=0 then we get for some x1 different than zero that |\int_{0}^{x_1}f(t)dt|&lt;2e and thus f(x1)=0.

i don't feel this proof is correct, but i understand intiutively why f(x) need to be equal zero.

 

[benorin]

i have another question:
prove that if f(x) is continuous and f(x)=\int_{0}^{x}f(t)dt then f(x) is zero.

what i did so far is:
for every e, there exists d, such that whenever |x-x1|<d, |f(x)-f(x1)|<e
|\int_{0}^{x}f(t)dt-\int_{0}^{x_1}f(t)dt|=|\int_{x_1}^{x}f(t)dt|&lt;e.
for every e, there exists d, such that whenever |x-x2|<d |f(x)-f(x2)|<e
we get that |x1-x2|<2d |f(x1)-f(x2)|<2e if x2=0 then we get for some x1 different than zero that |\int_{0}^{x_1}f(t)dt|&lt;2e and thus f(x1)=0.

i don't feel this proof is correct, but i understand intiutively why f(x) need to be equal zero.

If we add the additional requirement that f is differentiable, then we many employ the FTC to get f^{\prime}(x)=\frac{d}{dx}\int_{0}^{x}f(t)dt = f(x)

which is a differential equation whose only solution is f(x)=Ce^x but

f(x)=\int_{0}^{x}f(t)dt= C\int_{0}^{x}e^t \, dt = C(e^x-1)

so we must have Ce^x = C(e^x-1) so C=0 and so f(x)=0.

I know it's not the problem at hand, but perhaps it could be modified (or maybe you can prove that f(x) must be differentiable)?

 

[MathematicalPhysicist]

the only thing i tried so far by your advice is:
\forall\epsilon&gt;0, \exists\delta&gt;0, |x-x_0|=|h|&lt;\delta, |f(x)-f(x_0)|=|f(x_0+h)-f(x_0)|=|\int_{x_0}^{x_0+h}f(t)dt|&lt;\epsilon
and i need to show that lim[f(x0+h)-f(x0)]/h as h appraoches 0 exists, if we define:
g(h)=[f(x0+h)-f(x0)]/h then g(h) is continuous, but i need to show that's it continuous at h=0, which is ofcourse the hard part, perhaps dividing the first inequality with the epsilon by h?

 

[benorin]

Isn't f(x) the integral of something?

 

[MathematicalPhysicist]

yes, as i wrote :f(x)=\int_{0}^{x}f(t)dt

 

[benorin]

So f(x) is a primative (an antiderivative of something) and thus differentiable, right? now prove it.

 

[benorin]

From f(x)=\int_{0}^{x}f(t)dt clearly f(0)=0 (I'll use this later). On to differentiation:

f(x_0 +h)-f(x_0) = \int_{0}^{x_0+h}f(t)dt-\int_{0}^{x_0}f(t)dt = \int_{0}^{h}f(t)dt+\int_{h}^{x_0+h}f(t)dt-\int_{0}^{x_0}f(t)dt
=f(h)+\int_{0}^{x_0}f(t+h)dt-\int_{0}^{x_0}f(t)dt=f(h)+ \int_{0}^{x_0}[f(t+h)-f(t)]dt,

so we have

f^{\prime}(x_0)=\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}= \lim_{h\rightarrow 0}\frac{f(h)+ \int_{0}^{x_0}[f(t+h)-f(t)]dt}{h}
=\lim_{h\rightarrow 0}\frac{f(h)}{h}+ \lim_{h\rightarrow 0}\frac{1}{h}\int_{0}^{x_0}[f(t+h)-f(t)]dt = \lim_{h\rightarrow 0}\frac{f(0+h)-f(0)}{h}+ \lim_{h\rightarrow 0}\int_{0}^{x_0}\frac{f(t+h)-f(t)}{h}dt
=f^{\prime}(0)+ \lim_{h\rightarrow 0}\int_{0}^{x_0}\frac{f(t+h)-f(t)}{h}dt

I want to pass the limit through to the integrand to get \int_{0}^{x_0}f^{\prime}(t)dt and I think the continuity of f(x) justifies it, if so, then we have

f^{\prime}(x_0)=f^{\prime}(0)+\int_{0}^{x_0}f^{\prime}(t)dt = f^{\prime}(0)+f(x_0)-f(0)=f^{\prime}(0)+f(x_0)

which is another differential equation for f whose solution is f(x)=Ce^{x}-f^{\prime}(0).

CRAP! I realized that I assumed that f is differentiable after I passed the limit through to the integrand... hope some of this is useful. --Ben

 

[quasar987]

I don't have my books with me, but I think there is a thm that says that if f(x) is continuous, then the function F(x) defined by

F(x)=\int_0^x f(t)dt

is continuous, differentiable, and F'(x)=f(x).

In your problem, you are told that F(x)=f(x). So, f '(x)=f(x). The general solution of this is f(x)=Aexp(x). But f(0)=0, so A=0.

 

[0rthodontist]

f(x)=Aexp(x)+C

just A exp(x)

There must be a way to do this without depending on e^x. I'm wondering, how is it proved that the derivative of e^x is itself and that no other function has that property?

 

[quasar987]

just A exp(x)

I've made the appropriate correction to the post, thanks 0rthodontist.

I'm wondering, how is it proved that the derivative of e^x is itself and that no other function has that property?

The most general way* of defining the exponential function is through its power expansion. I.e. let's define the function exp(x) by

\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}

After showing that the series' radius of convergence is \infty, we can differentiate it term by term to find \frac{d}{dx}\exp(x) and see that it is exp(x) itself.

To prove exp(x) is the only function whose derivative is itself, "try" a series solution to the diff. equ. y'=y. It will come out that the only solution is the series with coefficients a_n=1/n!, which is our definiton of exp(x).*I say most general way because if we define exp(x) by say, the number e to the xth power (where e has been defined by say, the limit of (1+1/n)^n), then what is e^{ix+y}? What is a real number raised to a non-real power? But with the series definition, it is obvious what exp(ix+y) is.

 

[0rthodontist]

Then it depends on the whole machinery of infinite series.

 

[benorin]

I don't have my books with me, but I think there is a thm that says that if f(x) is continuous, then the function F(x) defined by

F(x)=\int_0^x f(t)dt

is continuous, differentiable, and F'(x)=f(x).

In your problem, you are told that F(x)=f(x). So, f '(x)=f(x). The general solution of this is f(x)=Aexp(x). But f(0)=0, so A=0.

This ODE has been derived and solved in posts #10 and #15, what is at hand is proving that f is differentiable based on the its continuity and the integral definition (I tried to do it the easy way, appearently an epsilon-delta type proof is prefered).

 

[shmoe]

This ODE has been derived and solved in posts #10 and #15, what is at hand is proving that f is differentiable based on the its continuity and the integral definition (I tried to do it the easy way, appearently an epsilon-delta type proof is prefered).

That's 'just' the fundamental theorem of calculus (one of them anyways), the mystery theorem quasar was recalling from his book- if f is continuous, then the F is differentiable (and it's derivative is f, etc). Proving this is now your analysis problem for the night (no peeking in a textbook).

 

[MathematicalPhysicist]

ok thank you for your help, it's just a matter of applying the fundamental theorem of calculus.

 

[HallsofIvy]

And remembering that the graph of an inverse function is just the graph of the original function reflected through the line y= x.

 

[MathematicalPhysicist]

i have another question on integral.
let f(x) be a Lipschitz-continuous on [0,1] that is, |f(x)-f(y)|<M|x-y|
for all x,y in the interval.
prove that:
|\int_{0}^{1}f(x)dx-\frac{\sum_{k=1}^{n}f(k/n)}{n}|&lt;M/2n
what i did is as follows:
the question asks me to proove that the integral exists, then if i use cauchy criterion for convergence, then the integral does exist.
if we let F_n=\sum_{k=1}^{n}f(k/n)*1/n , F_2n=\sum_{k=1}^{2n} f(k/2n)*1/2n then we get:
|F_{2n}-F_n|=|1/2n(f(1/2n)+f(1/n)+...+f(1))-1/n(f(1/n)+f(2/n)+...+f(1)|=1/2n|(f(1/2n)+f(3/2n)+...+f((n-1)/2n)-(f(1/n)+...+f(1))|&lt;=&lt;=1/2n[|f(1/2n)-f(1/n)|+...+|f((n-1)/2n)-f(1)|]&lt;(1/2n)M(1/2n+...+1/2n)=(1/2n)M/2&lt;M(1/2n)
in the last line i use lipshcitz condition, is this line of reasoning correct?

thanks in advance.

 

[MathematicalPhysicist]

has anyone given a look at my new problem here?
is my solution valid?