# A Integral with an inverse function limit

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1. Feb 20, 2017

Hello,

I have tried the integral below with Mathematica and it gives me the following solution:

$\frac{d}{dc}\int_{z^{-1}(c)}^{1} z(x)dx = -\frac{c}{z'(z^{-1}(c))}$

I am not quite sure where it gets it from...I think it can be separated and with differentiation the first part will be zero:

$\frac{d}{dc}\int_{z^{-1}(c)}^{1}z(x)dx=\frac{d}{dc}\int_{0}^{1} z(x)dx-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx=-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx$

By the formula $\frac{d}{dc}z^{-1}(c) = \frac{1}{z'(z^{-1}(c))}$, so it definitely has to play a role here, but how? and where does c come from? Do we need to somehow use chain rule here?

If the upper limit was just c it would be easy to take it with a Leibniz rule:

$-\frac{d}{dc}\int_{0}^{c} z(x)dx=z(c)$

However, it seems that when one of the limits is an inverse function, there is something different at work which I am not aware of...

2. Feb 23, 2017

### John Park

Let me struggle through this in my own way.
I find it useful to give familiar-looking names to things wherever possible.
So let's put
z-1(c) = y
and ∫z(x)dx = Z,
and I'll call the integral-derivative we have to evaluate F.

So we have to evaluate F = (d/dc)( Z(1) -Z(y) ),
which immediately becomes
-(d/dc)Z(y) = -(dZ(y)/dy). (dy/dc)
= z(y).(dy/dc)

But z(y) = z(z-1(c)) = c
So we have
F = c.(dy/dc) . . . . . . .(A)

Now z-1(c) = y by definition; so
c = z(y), and, differentiating,
dc/dy =dz/dy,
or if we invert,
dy/dc = 1/(dz/dy) = 1/z'(y)
= 1/z'( z-1(c) )

Substitute this for dy/dc in (A) and I think we're done.

3. Feb 26, 2017