If ∫₀ˣ f(t) dt = x + ∫ₓ¹ t f(t) dt, then the value of f(1) is
(a) 1/2
(b) 0
(c) 1
(d) -1/2

Q: Wait, why are we differentiating both sides? Isn't that changing the equation?

Great question! We aren't changing the equation; we are transforming it. Think of it like taking the derivative of both sides of $x^2 = 4$ to get $2x = 0$. We are performing an operation that preserves the equality. Since the original equation holds for all $x$, the derivatives of both sides must also be equal. This is a standard technique to convert an integral equation into a differential or algebraic equation.

Q: I always mess up the signs in the Leibniz rule. How do I stop doing that?

The sign error is the most common trap in JEE Advanced! Remember the rule: $\frac{d}{dx} \int_{a(x)}^{b(x)} g(t) dt = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$. Always write it out explicitly. If the lower limit is $x$, then $a'(x) = 1$, so you subtract $g(x) \cdot 1$. If you write the formula down every single time, you will stop making this mistake.

Q: Why did the integral $\int_x^1 t f(t) dt$ result in $-x f(x)$?

Let's look at the Leibniz rule again. The upper limit is $1$ (a constant), so its derivative is $0$. The lower limit is $x$, so its derivative is $1$. The rule says: (Function at upper limit $\times$ derivative of upper limit) - (Function at lower limit $\times$ derivative of lower limit). So, it becomes $(f(1) \cdot 0) - (x \cdot f(x) \cdot 1)$. That leaves us with $0 - x f(x)$, which is $-x f(x)$.

Question

If $\int_0^x f(t) dt = x + \int_x^1 t f(t) dt$, then the value of $f(1)$ is

Options:

(A) 1/2

(B) 0

(C) 1

(D) -1/2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first look at an expression like

$$\int_{0}^{x} f(t) \, dt = x + \int_{x}^{1} t f(t) \, dt$$

it is natural to feel a bit intimidated. You have an unknown function $f(t)$ trapped inside two different integrals, and the variable $x$ is dancing around in the limits.

In the world of JEE Advanced, these problems are not meant to be solved by brute force; they are meant to be solved by uncovering the hidden structure. Our goal is simple: find $f(1)$. To do that, we must liberate $f(x)$ from its integral prison.

### The Newton-Leibniz Rule

The key to this entire problem is the **Newton-Leibniz Rule**. Think of this rule as your master key. It allows us to differentiate an integral with respect to its limit.

The rule states:

$$\frac{d}{dx} \int_{a(x)}^{b(x)} g(t) \, dt = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$$

This is not just a formula; it is a geometric truth. It tells us how the area under a curve changes as we push the boundaries of the integration. If you memorize this, you have already won half the battle.

### The Differentiation

Now, let's apply this to our equation. We differentiate both sides with respect to $x$.

On the left side, we have $\frac{d}{dx} \int_{0}^{x} f(t) \, dt$. Applying the rule, we get $f(x) \cdot 1 - f(0) \cdot 0$, which simplifies beautifully to just $f(x)$.

Now, for the right side: $\frac{d}{dx} \left( x + \int_{x}^{1} t f(t) \, dt \right)$. The derivative of $x$ is $1$.

For the integral $\int_{x}^{1} t f(t) \, dt$, the upper limit is $1$ (a constant), so its derivative is $0$. The lower limit is $x$, so its derivative is $1$. Following the rule, we get $0 - (x \cdot f(x)) \cdot 1$.

This leaves us with $1 - x f(x)$. Be careful here! That negative sign is where many students lose marks.

### The Algebraic Resolution

We have now transformed our integral equation into a simple algebraic one:

$$f(x) = 1 - x f(x)$$

This is the moment of clarity. We have successfully extracted the function. Now, it is just a matter of rearranging the terms.

Add $x f(x)$ to both sides:

$$f(x) + x f(x) = 1$$

Factor out $f(x)$:

$$f(x)(1 + x) = 1$$

Finally, divide by $(1 + x)$ to get the general form:

$$f(x) = \frac{1}{1 + x}$$

### Final Calculation

We have found the function! The final step is trivial: evaluate at $x = 1$.

Substituting $x = 1$ into our result, we get:

$$f(1) = \frac{1}{1 + 1} = \frac{1}{2}$$

It is elegant, it is precise, and it is exactly the kind of problem that rewards a clear, methodical mind. Keep this technique in your toolkit; you will need it again. The final answer is **$1/2$**.

Animated Solution for Mathematics - Definite Integration: If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

Select Answer:

Visualized Solution

The Integral Equation

Newton-Leibniz Rule

Differentiating the LHS

Differentiating the RHS

Solving for $f (x)$

Evaluating $f (1)$

Key Takeaways

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Newton-Leibniz Rule

The Differentiation

The Algebraic Resolution

Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0x​f(t)dt=x+∫x1​tf(t)dt, then the value of f(1) is