Let f(x), x ≥ 0, be a non-negative continuous function, and let F(x) = ∫₀ˣ f(t) dt, x ≥ 0. If for so

Q: Wait, why can't we just divide by $F(x)$ to solve $f(x) \leq c F(x)$?

That is a very common temptation! If you divide by $F(x)$, you assume $F(x) > 0$. However, we don't know that yet. In fact, the whole point of the proof is to show that $F(x)$ is actually zero. Dividing by a potential zero is a mathematical sin that would lead you to a dead end.

Q: How do we know for sure that $F(0) = 0$?

Great question! Look at the definition $F(x) = \int_0^x f(t) dt$. If you plug in $x=0$, you get $F(0) = \int_0^0 f(t) dt$. By the definition of a definite integral, the area from a point to itself is always zero. It is the anchor point for our entire proof.

Q: Why does $F(x) \leq 0$ and $F(x) \geq 0$ force $F(x) = 0$?

This is the 'sandwich' or 'squeeze' logic. If a value is simultaneously less than or equal to zero and greater than or equal to zero, the only number that satisfies both conditions is zero. It is the only place where these two inequalities meet.

Question

Let $f(x), x \geq 0$, be a non-negative continuous function, and let $F(x) = \int_0^x f(t) dt, x \geq 0$. If for some $c > 0, f(x) \leq c F(x)$ for all $x \geq 0$, then show that $f(x) = 0$ for all $x \geq 0$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plain. You have a function, $f(x)$, which represents the height of a landscape.

We are told this landscape is non-negative, meaning it never dips below the ground level of the $x$-axis.

We introduce a new function, $F(x)$, defined as the area under this landscape from the origin to some point $x$:

$$F(x) = \int_0^x f(t) dt$$

This is our area function, a memory of the past that accumulates the total area as we walk forward.

### The Bridge Between Area and Curve

How are the area and the curve related? This is the heart of the Fundamental Theorem of Calculus.

The rate at which the area grows is exactly the height of the curve at that point. Mathematically, we write this as:

$$F'(x) = f(x)$$

This is our logic bridge. It allows us to translate a statement about the curve into a statement about the area function.

### The Differential Inequality

The problem gives us a fascinating constraint: $f(x) \leq c F(x)$. This tells us that the height of the landscape is limited by the area accumulated so far.

If the area is small, the landscape must be low. If the area is large, the landscape can be higher.

Let's substitute our bridge into this constraint:

$$F'(x) \leq c F(x)$$

Rearranging this, we get $F'(x) - c F(x) \leq 0$. This looks like a linear differential equation, but with an inequality.

This is where we pull a powerful tool from our JEE toolkit: the **Integrating Factor**. We want to turn the left side into the derivative of a product.

We multiply the entire inequality by $e^{-cx}$. Because $e^{-cx}$ is always strictly positive, our inequality sign remains unchanged:

$$e^{-cx} F'(x) - c e^{-cx} F(x) \leq 0$$

### The Elegance of the Product Rule

Look closely at the left-hand side. It is the exact expansion of the product rule for derivatives!

We can condense this into:

$$\frac{d}{dx} (e^{-cx} F(x)) \leq 0$$

This is the moment of clarity. We have a function whose derivative is always non-positive.

This means the function $e^{-cx} F(x)$ must be non-increasing.

### The Final Contradiction

Now, let's integrate both sides from $0$ to $x$. The integral of the derivative gives us back the original function:

$$\left[ e^{-ct} F(t) \right]_0^x \leq \int_0^x 0 dt$$

Evaluating this, we get $e^{-cx} F(x) - e^0 F(0) \leq 0$. Since $F(0) = 0$, this simplifies beautifully to:

$$e^{-cx} F(x) \leq 0$$

Since $e^{-cx}$ is always positive, this forces $F(x) \leq 0$. But wait!

We know that $f(x) \geq 0$, so the area under the curve must be non-negative, meaning $F(x) \geq 0$.

We are trapped between $F(x) \leq 0$ and $F(x) \geq 0$. The only way out is **$F(x) = 0$**.

If the area function is constantly zero, its derivative, $f(x)$, must also be zero. The landscape is not just low; it is a perfectly flat line on the $x$-axis. We have proven it!

Animated Solution for Mathematics - Differential Equations: Let $f (x), x \geq 0$ , be a non-negative continuous function, and let $F (x) = \int_{0}^{x} f (t) d t, x \geq 0$ . If for some $c > 0, f (x) \leq c F (x)$ for all $x \geq 0$ , then show that $f (x) = 0$ for all $x \geq 0$ .

Visualized Solution

Understanding the Given Functions

The Area Function $F (x)$

The Fundamental Theorem of Calculus

Setting Up the Inequality

Forming the Differential Inequality

The Integrating Factor

Multiplying by the Integrating Factor

Condensing with the Product Rule

Integrating the Inequality

Evaluating the Definite Integral

Using the Initial Condition

Analyzing the Sign of $F (x)$

The Sandwich Logic

Concluding $F (x) = 0$

Final Conclusion for $f (x)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Bridge Between Area and Curve

The Differential Inequality

The Elegance of the Product Rule

The Final Contradiction

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