Let f: R - R be a continuous function which satisfies f(x) = ∫₀ˣ f(t) dt. Then the value of f(ln 5)

Q: Wait, why can't I just guess that $f(x) = e^x$? That satisfies the derivative part!

That is a brilliant observation! $f(x) = e^x$ is indeed a solution to the differential equation $f'(x) = f(x)$. However, a differential equation is only half the story. You must also satisfy the original integral equation. If you plug $e^x$ into $\int_{0}^{x} e^t dt$, you get $e^x - 1$, which is not equal to $e^x$. The initial condition $f(0)=0$ acts as a filter that rejects $e^x$ and keeps only the trivial solution.

Q: Why does the integral from 0 to 0 always equal 0?

Think of the integral as the area under a curve. If the width of the interval is zero (from $x=0$ to $x=0$), there is no 'base' for your rectangle, so the area must be zero. Mathematically, $\int_{a}^{a} f(t) dt = F(a) - F(a) = 0$.

Q: Is it possible for a function to be its own derivative and still be zero?

Yes! The zero function $f(x) = 0$ is the only function that is its own derivative and passes through the origin. It is the 'trivial' solution, but in mathematics, trivial doesn't mean unimportant—it means it's the foundation upon which everything else rests.

Q: Do I always have to differentiate when I see an integral equation?

Not always, but it is a powerful 'weapon' in your JEE toolkit. Whenever you see a variable in the limit of an integral, differentiating is usually the fastest way to turn a complex integral equation into a manageable differential equation.

Q: What if the integral was from 1 to x instead of 0 to x?

That would change everything! If the lower limit were 1, then $f(1) = \int_{1}^{1} f(t) dt = 0$. Your initial condition would be $f(1)=0$ instead of $f(0)=0$. You would solve $f(x) = Ce^x$ using $f(1) = Ce^1 = 0$, which still leads to $C=0$. The result would remain the same, but the logic path would shift!

Question

Let $f: R 	o R$ be a continuous function which satisfies $f(x) = \int_{0}^{x} f(t) dt$. Then the value of $f(\ln 5)$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a mirror, but instead of seeing your own reflection, you see a mathematical equation staring back at you: $f(x) = \int_{0}^{x} f(t) dt$.

This is not just any equation; it is a self-referential loop. The function $f(x)$ is defined by its own accumulated history—the area it has swept out from the origin to the point $x$.

### The Power of Differentiation

When we see an integral with a variable limit, our instinct should be to 'unlock' it. We want to strip away the integral sign to see the function in its raw, differential form.

To do this, we invoke the mighty Newton-Leibniz Rule. By differentiating both sides with respect to $x$, we are essentially asking: 'How does the area under this curve change as we push the boundary $x$ slightly further?'

On the left, the derivative of $f(x)$ is simply $f'(x)$. On the right, the derivative of $\int_{0}^{x} f(t) dt$ is, by the Fundamental Theorem of Calculus, just $f(x)$.

Suddenly, the complexity vanishes, leaving us with a beautiful, elegant differential equation:

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

### The Soul of the Differential Equation

This equation, $f'(x) = f(x)$, is the heartbeat of calculus. It describes a function whose rate of growth is perfectly proportional to its current value.

We know the solution to this: the exponential family, $f(x) = Ce^x$. But wait—before we celebrate, we must remember that we are not just solving a differential equation; we are solving the original integral equation.

We have a constant $C$ that needs to be determined.

### The Gatekeeper: Initial Conditions

In physics and mathematics, the initial condition is the gatekeeper of truth. Let us return to our original equation: $f(x) = \int_{0}^{x} f(t) dt$.

What happens if we set $x = 0$? The right side becomes the integral from $0$ to $0$, which is geometrically and algebraically zero. Thus, we discover the hidden constraint: $f(0) = 0$.

Now, let us apply this to our general solution, $f(x) = Ce^x$. If we plug in $x = 0$, we get $f(0) = C \cdot e^0 = C$.

Since we already established that $f(0) = 0$, it forces the constant $C$ to be exactly $0$.

### The Trivial Truth

It might feel anticlimactic to find that $C = 0$, leading us to the conclusion that $f(x) = 0$ for all $x$. You might ask, 'Is that it?'

But look at the beauty of it! The only function that can satisfy the condition of being equal to its own integral starting from the origin is the zero function itself. It is a perfect, balanced state of nothingness.

When we are asked to find $f(\ln 5)$, we simply look at our result. Since $f(x)$ is identically zero for every real number, **$f(\ln 5) = 0$**.

We have navigated the integral, differentiated the mystery, applied the boundary conditions, and arrived at the truth. Keep this logic in your heart—whenever you face a complex integral equation, look for the derivative, find the initial condition, and let the math reveal the answer.

Animated Solution for Mathematics - Differential Equations: Let $f : R \to R$ be a continuous function which satisfies $f (x) = \int_{0}^{x} f (t) d t$ . Then the value of $f (ln 5)$ is

Enter Numerical Value:

Visualized Solution

The Integral Equation

Geometric Interpretation

Differentiating the Integral

Applying Leibniz Rule

The Differential Equation

Rearranging the Equation

Integrating Both Sides

The General Solution

Finding the Initial Condition

Evaluating $f (0)$

Solving for $C$

The Trivial Solution

Final Evaluation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Power of Differentiation

The Soul of the Differential Equation

The Gatekeeper

The Trivial Truth

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a continuous function which satisfies f(x)=∫0x​f(t)dt. Then the value of f(ln5) is