If f(x) is continuous and positive, prove that

If $f(x)$ is continuous and positive, prove that the function $$ \varphi(x) = \frac{\int_0^x t f(t) \, dt}{\int_0^x f(t) \, dt} $$ is increasing on the interval $[0, \infty)$.

This function $\varphi(x)$ represents the weighted average (or centroid) of the interval $[0, x]$ with weight function $f(t)$. We will prove it's increasing by showing $\varphi'(x) \geq 0$.

Setup and interpretation

The function $\varphi(x)$ can be interpreted as:

$$\varphi(x) = \frac{\text{first moment}}{\text{total mass}} = \text{weighted average of } t \text{ on } [0, x]$$

where the weight at point $t$ is $f(t)$.

Step 1: Find the derivative $\varphi'(x)$

We have a quotient, so we'll use the quotient rule. Let:

$$u(x) = \int_0^x t f(t) \, dt \quad \text{and} \quad v(x) = \int_0^x f(t) \, dt$$

By the Fundamental Theorem of Calculus:

$$u'(x) = xf(x) \quad \text{and} \quad v'(x) = f(x)$$

Using the quotient rule:

$$\varphi'(x) = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{[v(x)]^2}$$

$$= \frac{xf(x) \cdot \int_0^x f(t) \, dt - \int_0^x t f(t) \, dt \cdot f(x)}{\left[\int_0^x f(t) \, dt\right]^2}$$

Factor out $f(x)$ from the numerator:

$$= \frac{f(x) \left[x \int_0^x f(t) \, dt - \int_0^x t f(t) \, dt\right]}{\left[\int_0^x f(t) \, dt\right]^2}$$

$$= \frac{f(x)}{\left[\int_0^x f(t) \, dt\right]^2} \left[x \int_0^x f(t) \, dt - \int_0^x t f(t) \, dt\right]$$

Step 2: Simplify the bracket expression

The key expression is:

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

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

Step 3: Analyze the sign of $\varphi'(x)$

We have:

$$\varphi'(x) = \frac{f(x)}{\left[\int_0^x f(t) \, dt\right]^2} \int_0^x (x-t)f(t) \, dt$$

Let's examine each factor:

First factor: $f(x) > 0$ by assumption ✓

Second factor: $\left[\int_0^x f(t) \, dt\right]^2 > 0$ (positive since $f > 0$ and we're integrating over a non-trivial interval for $x > 0$)

Third factor: $\int_0^x (x-t)f(t) \, dt$

For $t \in [0, x]$, we have $x - t \geq 0$ (with equality only at $t = x$).

Since $f(t) > 0$ for all $t$, the integrand $(x-t)f(t) \geq 0$ for all $t \in [0, x]$.

Moreover, $(x-t)f(t) > 0$ for $t \in [0, x)$ (strictly positive except at the single point $t = x$).

Therefore:

$$\int_0^x (x-t)f(t) \, dt > 0 \quad \text{for } x > 0$$

Step 4: Conclusion

Since all three factors are positive for $x > 0$:

$$\varphi'(x) > 0 \quad \text{for all } x > 0$$

This proves that $\varphi(x)$ is strictly increasing on $(0, \infty)$.

At $x = 0$, the function $\varphi(x)$ is not defined (both numerator and denominator are 0), but we can find its limit.

Step 5: Behavior at $x = 0$

Using L'Hôpital's rule:

$$\lim_{x \to 0^+} \varphi(x) = \lim_{x \to 0^+} \frac{\int_0^x t f(t) \, dt}{\int_0^x f(t) \, dt}$$

Both numerator and denominator approach 0, so applying L'Hôpital's rule:

$$= \lim_{x \to 0^+} \frac{xf(x)}{f(x)} = \lim_{x \to 0^+} x = 0$$

So if we define $\varphi(0) = 0$, the function becomes continuous at $x = 0$ and increasing on $[0, \infty)$.

Final Answer

$$\boxed{\varphi'(x) = \frac{f(x) \int_0^x (x-t)f(t) \, dt}{\left[\int_0^x f(t) \, dt\right]^2} > 0 \quad \text{for } x > 0}$$

Therefore, $\varphi(x)$ is strictly increasing on $(0, \infty)$.

Geometric and physical interpretation

As a weighted average:

The function $\varphi(x)$ represents the weighted average position on $[0, x]$ with weight density $f(t)$.

As $x$ increases, we're adding mass at position $x$ (which is to the right of the current average). This necessarily shifts the weighted average to the right, hence $\varphi(x)$ increases.

As a center of mass:

If we think of $f(t)$ as a mass density along a rod from $0$ to $x$, then:

• $\int_0^x f(t) \, dt$ = total mass
• $\int_0^x t f(t) \, dt$ = first moment about the origin
• $\varphi(x)$ = center of mass position

As the rod extends from length $x$ to length $x + dx$, we add mass at position $x$, which is to the right of the current center of mass (for $x > \varphi(x)$, which we'll verify below). This shifts the center of mass to the right.

Verification that $x > \varphi(x)$ for $x > 0$

We need to show:

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

Multiplying both sides by $\int_0^x f(t) \, dt > 0$:

$$x \int_0^x f(t) \, dt > \int_0^x t f(t) \, dt$$

$$\int_0^x x f(t) \, dt > \int_0^x t f(t) \, dt$$

$$\int_0^x (x - t) f(t) \, dt > 0$$

This is true since $x - t > 0$ for $t \in [0, x)$ and $f(t) > 0$. ✓

Therefore, the current position $x$ is always to the right of the weighted average $\varphi(x)$, confirming our intuition.

Example: Uniform weight $f(t) = 1$

If $f(t) = 1$ (uniform weighting), then:

$$\varphi(x) = \frac{\int_0^x t \, dt}{\int_0^x 1 \, dt} = \frac{\frac{x^2}{2}}{x} = \frac{x}{2}$$

This is the midpoint of $[0, x]$, and indeed $\varphi(x) = \frac{x}{2}$ is increasing with $\varphi'(x) = \frac{1}{2} > 0$. ✓

Example: Exponential weight $f(t) = e^t$

$$\varphi(x) = \frac{\int_0^x t e^t \, dt}{\int_0^x e^t \, dt}$$

Using integration by parts for the numerator:

$$\int_0^x t e^t \, dt = [te^t]_0^x - \int_0^x e^t \, dt = xe^x - (e^x - 1) = xe^x - e^x + 1$$

The denominator:

$$\int_0^x e^t \, dt = e^x - 1$$

Therefore:

$$\varphi(x) = \frac{xe^x - e^x + 1}{e^x - 1} = \frac{e^x(x-1) + 1}{e^x - 1}$$

We can verify this is increasing, though the algebra is more complex.

Summary

1. We computed the derivative using the quotient rule: $$ \varphi'(x) = \frac{f(x) \int_0^x (x-t)f(t) \, dt}{\left[\int_0^x f(t) \, dt\right]^2} $$
2. Since $f(x) > 0$, $\int_0^x f(t) \, dt > 0$, and $(x-t)f(t) \geq 0$ with strict inequality on $[0, x)$, we have: $$ \varphi'(x) > 0 \quad \text{for all } x > 0 $$
3. This proves $\varphi(x)$ is strictly increasing on $(0, \infty)$
4. The result has a natural interpretation: the weighted average increases as we extend the interval and add mass to the right

Key lessons:

• The quotient rule combined with the Fundamental Theorem of Calculus is powerful for analyzing ratios of integrals
• Weighted averages have monotonicity properties that reflect adding mass at the boundary
• The key insight is recognizing $\int_0^x (x-t)f(t) \, dt > 0$, which shows the current position is to the right of the average
• Physical intuition (center of mass) can guide mathematical proofs