# Integrating with respect to y

1. Nov 26, 2013

### MathewsMD

Question: Suppose f is continuous, f(0) = 0, f(1) =1, f'(x) > 0, and ∫01f(x)dx = 1/3. Find the value of the integral of f-1(y)dy

One solution is to assess the function as if it were a function of y. I understand that method and have arrived at the answer.

But I am curious to see if there is another solution since I have been unable to come up with another method besides just looking at the graph visually after I rotate it. If there is a more general answer to assessing the integral of inverse functions, that would be great if you could provide an explanation as well!

Also, if you were asked to solve this: ∫01 d/dx f-1(y)dy, is it possible with the information given above alone? If not, what additional information is necessary?

Also, are there any general rules when integrating inverse functions?

Thank you so much!

2. Nov 26, 2013

### Staff: Mentor

Presumably the second integral is
$$\int_0^1 f^{-1}(y)dy$$
I get a value of 2/3 for this integral.

I don't understand what you're saying.
If y = f(x), and f'(x) > 0, then f is increasing. This also implies that f is one-to-one, so has an inverse that is a function. This means that the equation y = f(x) can be written as x = f-1(y), which is an equivalent equation. IOW, any pair (x, y) that satisfies y = f(x) also satisfies x = f-1(y).

From the given information, we can sketch a reasonable graph of f. The curve has to be concave up, since the value of the given integral is 1/3, which is less than half of the area of the rectangle whose opposite corners are at the origin and (1, 1).

This integral--
$$\int_0^1 f^{-1}(y)dy = \int_0^1 x dy$$
-- represents the area of the region bounded below by the graph of f, to the left by the y-axis, and above by the line y = 1. The typical area element of this integral is a thin horizontal strip that is x in length (= f-1(y)) by Δy in width.
There's no need to rotate anything, if you understand how a function and its inverse are related.
Sure. Since x = f-1(y), the integrand can be simplified to d/dx(x), or 1, integrated with respect to y.

3. Nov 26, 2013

### brmath

Here's what I've worked out:

$\int_0^1 f^{-1}(f(x)) \cdot f'(x)dx = \int_{f(0)}^{f(1)} f^{-1}(u)du = \int_0^1 f^{-1}(u)du$ from substituting u = f(x). However the first integral also gives

$\int_0^1 f^{-1}(f(x)) \cdot f'(x)dx = xf(x)\Bigg|_0^1 - \int_0^1f(x)dx$ noting that $f^{-1}(f(x))$ = x and using integration by parts.

This last simplifies down to 1 - 1/3 =2/3. Agreeably, this is what Mark44 gets.

While this is probably equivalent to what you did with the y, I think this method states the matter pretty generally.