Animated Solution for Mathematics - Definite Integration: If $f:R\to R$ is a differentiable function and $f(2)=6$, then ${\lim }_{x\to 2}\frac{{\int }_6^{f(x)}2tdt}{x-2}$ is :-

My dear student, welcome to the arena of JEE Advanced calculus. Today, we confront a problem that often intimidates students because it blends two distinct worlds: the world of limits and the world of definite integrals.

When you see the expression:

Your first instinct might be to panic. But let us pause. In mathematics, as in life, the first step is always to assess the situation.

Let us test the waters by substituting $x=2$ directly into the expression. The denominator becomes $2-2=0$.

Now, look at the numerator: ${\int }_6^{f(2)}2tdt$. We are given that $f(2)=6$, so the integral becomes ${\int }_6^{6}2tdt$.

The area under a curve from a point to itself is, by definition, zero. We have arrived at a $\frac{0}{0}$ indeterminate form. This is not a wall; it is a gateway. It tells us that the limit exists and that we have the green light to use our most powerful tools.

The Newton-Leibniz Formula

Since we have a $\frac{0}{0}$ form, our immediate reflex should be L'Hopital's Rule. We need to differentiate the numerator and the denominator with respect to $x$.

The denominator is trivial: $\frac{d}{dx}(x-2)=1$. But the numerator? That is where the magic happens.

We need to differentiate an integral where the variable $x$ is trapped in the upper limit: $\frac{d}{dx}{\int }_6^{f(x)}2tdt$. This is where we summon the Newton-Leibniz formula.

It states that:

Think of this as the Chain Rule for integrals. We are essentially asking: how does the area under the curve change as the boundary $f(x)$ moves? We replace $t$ with the upper limit $f(x)$ and multiply by the rate at which that limit is changing, which is $f^{\prime }(x)$.

Let us apply this with precision. Our integrand is $g(t)=2t$. The upper limit is $b(x)=f(x)$, and its derivative is $b^{\prime }(x)=f^{\prime }(x)$.

The lower limit is $a(x)=6$, a constant, so $a^{\prime }(x)=0$. Plugging this into our formula, the numerator's derivative becomes:

The second term vanishes into thin air, leaving us with $2f(x)f^{\prime }(x)$. Now, we return to our limit. We have transformed the terrifying integral expression into the simple limit:

We are at the finish line. As $x$ approaches $2$, $f(x)$ approaches $f(2)$, which is $6$. The expression becomes $2\cdot f(2)\cdot f^{\prime }(2)$.

Substituting the known value $f(2)=6$, we get $2\cdot 6\cdot f^{\prime }(2)$, which simplifies beautifully to $12f^{\prime }(2)$.

Do you see the elegance? We started with an integral that seemed to require complex evaluation, and through the power of the Leibniz Rule, we reduced it to a simple algebraic product.

This is the essence of JEE Advanced: identifying the right tool, applying it with discipline, and watching the complexity collapse into a simple, elegant answer. Keep practicing this, and you will find that no limit can stand in your way.