Solving for the Integral of the Inverse Function f-1(y)dy

[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!

 

[Mark44]

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

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

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.

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.

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.

There's no need to rotate anything, if you understand how a function and its inverse are related.

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?

Sure. Since x = f^-1(y), the integrand can be simplified to d/dx(x), or 1, integrated with respect to y.

If not, what additional information is necessary?

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

Thank you so much!

 

[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.