Let f: R - R be a differentiable function having f(2) = 6, f'(2) = 1/48. Then lim_{x-2} ∫₆ᶠ⁽ˣ⁾ 4t³

Q: Wait, why do we use L'Hopital's Rule here? Can't we just evaluate the integral first?

Great question! You could technically evaluate the integral $\int_6^{f(x)} 4t^3 dt$ to get $[t^4]_6^{f(x)} = (f(x))^4 - 6^4$. Then the limit becomes $\lim_{x \to 2} \frac{(f(x))^4 - 6^4}{x-2}$. This is the definition of the derivative of $h(x) = (f(x))^4$ at $x=2$. Using the chain rule, that is $4(f(2))^3 \cdot f'(2)$. Both methods lead to the same beautiful result!

Q: Why is the derivative of the lower limit 6 equal to zero?

Remember that the derivative measures the rate of change. Since 6 is a constant, it doesn't change as $x$ changes. Therefore, its rate of change is zero. In the Leibniz formula, the term involving the lower limit vanishes because the lower limit is a fixed constant.

Q: Is it always safe to assume the limit is $0/0$?

In JEE problems, always check the form first! If the numerator were a non-zero constant, the limit would be infinity. Always verify the indeterminate form before applying L'Hopital's Rule to ensure your steps are logically sound.

Q: What if $f(x)$ was in the denominator instead of the numerator?

If the integral were in the denominator, you would still have a $0/0$ form. You would apply L'Hopital's Rule to both the numerator and the denominator, differentiating the integral using the Leibniz rule just as we did here. The process remains consistent!

Q: I keep getting confused with the chain rule inside the integral. Any tips?

Think of the integral as a function of its upper limit. If you have $\int_a^{f(x)} g(t) dt$, you are essentially composing the integral function with $f(x)$. The chain rule tells us to differentiate the outer function (the integral) and multiply by the derivative of the inner function ($f'(x)$). The fundamental theorem of calculus handles the 'outer' part, and $f'(x)$ handles the 'inner' part.

Question

Let $f: R 	o R$ be a differentiable function having $f(2) = 6, f'(2) = \frac{1}{48}$. Then $\lim_{x 	o 2} \int_6^{f(x)} \frac{4t^3}{x-2} dt$ equals

Options:

(A) $24$

(B) $36$

(C) $12$

(D) $18$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, might seem like a daunting wall of symbols. We have a function $f(x)$, an integral with a variable upper limit, and a limit approaching a specific point.

In the world of advanced mathematics, complexity is often just a mask for elegance. Let us peel back that mask together.

### The Indeterminate Gateway

Before we rush into calculations, we must understand the landscape. We are given the limit:

$$\lim_{x 	o 2} \frac{\int_6^{f(x)} 4t^3 dt}{x-2}$$

The first rule of limits is: always test the waters. As $x 	o 2$, the denominator $x-2$ clearly heads toward zero.

Regarding the numerator, as $x 	o 2$, $f(x) 	o f(2) = 6$. Thus, the integral becomes:

$$\int_6^6 4t^3 dt = 0$$

We have arrived at the classic $\frac{0}{0}$ indeterminate form. This is not a dead end; it is an invitation to use **L'Hopital's Rule**.

### The Power of Leibniz

To apply L'Hopital's Rule, we must differentiate the numerator and the denominator with respect to $x$. The derivative of the denominator $x-2$ is simply $1$.

For the numerator, we invoke the **Newton-Leibniz formula**. We are looking at:

$$\frac{d}{dx} \int_6^{f(x)} 4t^3 dt$$

The rule states that for $H(x) = \int_a^{f(x)} g(t) dt$, the derivative is $g(f(x)) \cdot f'(x)$. Applying this to our expression, we get:

$$4(f(x))^3 \cdot f'(x)$$

This is the heart of the problem—the moment where the calculus simplifies into a beautiful, manageable expression.

### The Final Synthesis

Now, our limit has transformed from a terrifying integral into a clean, algebraic expression:

$$\lim_{x 	o 2} 4(f(x))^3 f'(x)$$

We are given the values $f(2) = 6$ and $f'(2) = \frac{1}{48}$. Substituting these values as $x 	o 2$, we obtain:

$$4 \cdot (6)^3 \cdot \frac{1}{48}$$

Calculating the result:

$$4 \cdot 216 \cdot \frac{1}{48} = \frac{864}{48} = 18$$

The final answer is **18**.

### The Takeaway

We started with an intimidating integral and ended with a simple integer. This is the beauty of the JEE Advanced curriculum.

When you see a limit involving an integral, do not panic. Check for the indeterminate form, use the Leibniz rule to peel away the integral, and let the chain rule guide you to the answer. You have the tools, the logic, and the experience to conquer these challenges.

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be a differentiable function having $f (2) = 6, f^{'} (2) = \frac{1}{48}$ . Then $lim_{x \to 2} \int_{6}^{f (x)} \frac{4 t ^{3}}{x - 2} d t$ equals

Select Answer:

Visualized Solution

Analyze the Limit Form

Apply L'Hopital's Rule

Newton-Leibniz Formula

Differentiating the Numerator

Differentiating the Denominator

Substitute the Values

Final Calculation

The Way Forward

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Indeterminate Gateway

The Power of Leibniz

The Final Synthesis

The Takeaway

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