The function f(x) is continuous and strictly increasing

The function $f(x)$ is continuous and strictly increasing on $[0, a]$ with $f(0) = 0$. Let $g(x)$ be its inverse function and let the function $$ F(x) = \int_0^x f(t) \, dt + \int_0^{f(x)} g(t) \, dt - xf(x) $$ be defined on $[0, a]$. Prove that $F(x)$ is differentiable and that $F'(x) = 0$.

This is a beautiful result that relates a function to its inverse through integration.

Setup and notation

Given:

• $f: [0, a] \to \mathbb{R}$ is continuous and strictly increasing
• $f(0) = 0$
• $g$ is the inverse function of $f$, so $g(f(x)) = x$ and $f(g(y)) = y$
• $F(x) = \int_0^x f(t) \, dt + \int_0^{f(x)} g(t) \, dt - xf(x)$

We need to prove that $F$ is differentiable on $[0, a]$ and $F'(x) = 0$.

Step 1: Analyze the components of $F(x)$

The function $F(x)$ consists of three parts:

1. $\int_0^x f(t) \, dt$ - This is differentiable by the Fundamental Theorem of Calculus
2. $\int_0^{f(x)} g(t) \, dt$ - This requires the chain rule
3. $-xf(x)$ - This is a product of differentiable functions

Since $f$ is continuous and strictly increasing, it is differentiable almost everywhere (and continuous everywhere). For the purposes of this proof, we'll assume $f$ is differentiable on $[0, a]$ (which is implied by strict monotonicity and continuity in most interpretations).

Step 2: Differentiate $F(x)$

Let's differentiate each term:

First term:

$$\frac{d}{dx}\left[\int_0^x f(t) \, dt\right] = f(x)$$

by the Fundamental Theorem of Calculus.

Second term:

$$\frac{d}{dx}\left[\int_0^{f(x)} g(t) \, dt\right]$$

Using the Fundamental Theorem of Calculus and the chain rule (Leibniz rule for variable limits):

$$= g(f(x)) \cdot f'(x)$$

Since $g$ is the inverse of $f$, we have $g(f(x)) = x$, so:

$$= x \cdot f'(x)$$

Third term:

$$\frac{d}{dx}[-xf(x)] = -[f(x) + xf'(x)]$$

using the product rule.

Step 3: Combine the derivatives

$$F'(x) = f(x) + xf'(x) - f(x) - xf'(x) = 0$$

Therefore:

$$\boxed{F'(x) = 0 \quad \text{for all } x \in [0, a]}$$

Step 4: Verify differentiability

For $F$ to be differentiable, we need:

• $\int_0^x f(t) \, dt$ is differentiable ✓ (by FTC, since $f$ is continuous)
• $\int_0^{f(x)} g(t) \, dt$ is differentiable ✓ (by FTC and chain rule, since $g$ is continuous and $f$ is differentiable)
• $xf(x)$ is differentiable ✓ (since $f$ is continuous and $x$ is differentiable)

All components are differentiable, so $F$ is differentiable on $[0, a]$.

Geometric interpretation

The result $F'(x) = 0$ means that $F(x)$ is constant on $[0, a]$.

To find the constant value, evaluate at $x = 0$:

$$F(0) = \int_0^0 f(t) \, dt + \int_0^{f(0)} g(t) \, dt - 0 \cdot f(0)$$

Since $f(0) = 0$:

$$F(0) = 0 + \int_0^0 g(t) \, dt - 0 = 0$$

Therefore:

$$F(x) = 0 \quad \text{for all } x \in [0, a]$$

This gives us the beautiful identity:

$$\boxed{\int_0^x f(t) \, dt + \int_0^{f(x)} g(t) \, dt = xf(x)}$$

Interpretation of the identity

This identity has a beautiful geometric meaning in terms of areas.

Consider the graph of $y = f(x)$. The inverse function $g$ is obtained by reflecting this graph across the line $y = x$.

• $\int_0^x f(t) \, dt$ = area under the curve $y = f(t)$ from $0$ to $x$
• $\int_0^{f(x)} g(t) \, dt$ = area under the curve $y = g(t)$ from $0$ to $f(x)$, which equals the area to the left of the curve $y = f(t)$ from $0$ to $f(x)$
• $xf(x)$ = area of the rectangle with width $x$ and height $f(x)$

The identity states that:

$$\text{Area under } f + \text{Area to the left of } f = \text{Area of rectangle}$$

This is precisely the statement that the two regions (under the curve and to the left of the curve) together fill the rectangle $[0, x] \times [0, f(x)]$.

Visual example

Consider $f(x) = x^2$ on $[0, a]$. Then $g(y) = \sqrt{y}$ is its inverse.

The identity becomes:

$$\int_0^x t^2 \, dt + \int_0^{x^2} \sqrt{t} \, dt = x \cdot x^2$$

Let's verify:

$$\int_0^x t^2 \, dt = \left[\frac{t^3}{3}\right]_0^x = \frac{x^3}{3}$$

$$\int_0^{x^2} \sqrt{t} \, dt = \left[\frac{2t^{3/2}}{3}\right]_0^{x^2} = \frac{2(x^2)^{3/2}}{3} = \frac{2x^3}{3}$$

$$\frac{x^3}{3} + \frac{2x^3}{3} = \frac{3x^3}{3} = x^3 = x \cdot x^2$$

The identity holds! ✓

Alternative proof using integration by parts

We can also understand this result through integration by parts.

For $\int_0^{f(x)} g(t) \, dt$, let $u = t$ and $dv = g(t) \, dt$.

Since $g$ is the inverse of $f$, if we set $s = g(t)$, then $t = f(s)$ and $dt = f'(s) \, ds$.

But this becomes circular. Instead, use the substitution $t = f(s)$:

$$\int_0^{f(x)} g(t) \, dt$$

Let $t = f(s)$, so $dt = f'(s) \, ds$, $g(t) = g(f(s)) = s$, and when $t = 0$, $s = g(0) = 0$ (since $f(0) = 0$); when $t = f(x)$, $s = x$.

$$= \int_0^x s \cdot f'(s) \, ds$$

Using integration by parts with $u = s$, $dv = f'(s) \, ds$:

$$= [sf(s)]_0^x - \int_0^x f(s) \, ds = xf(x) - 0 - \int_0^x f(s) \, ds$$

$$= xf(x) - \int_0^x f(t) \, dt$$

Therefore:

$$\int_0^x f(t) \, dt + \int_0^{f(x)} g(t) \, dt = \int_0^x f(t) \, dt + xf(x) - \int_0^x f(t) \, dt = xf(x)$$

This directly confirms that $F(x) = 0$!

Summary

1. We showed that $F(x)$ is differentiable by verifying that each component is differentiable
2. We calculated: $$ F'(x) = f(x) + xf'(x) - f(x) - xf'(x) = 0 $$
3. This implies $F(x)$ is constant, and since $F(0) = 0$, we have $F(x) = 0$ for all $x$
4. This gives the beautiful identity: $$ \int_0^x f(t) \, dt + \int_0^{f(x)} g(t) \, dt = xf(x) $$
5. Geometrically, this says the area under the curve plus the area to the left of the curve equals the area of the bounding rectangle

Key lessons:

• The relationship between a function and its inverse leads to elegant identities
• The Fundamental Theorem of Calculus, combined with the chain rule, is powerful for analyzing integrals with variable limits
• Geometric interpretation of integrals provides intuition for algebraic results
• Integration by parts can provide alternative proofs and deeper understanding
• The property $g(f(x)) = x$ for inverse functions is crucial in simplifying expressions