Animated Solution for Mathematics - Differentiation: If $f(x)={\int }_0^{x^2}{e}^{\sqrt{t}}(t-2)(t-3)dt$ for all $x\in (0,\infty )$, then

Unveiling the Hidden Geometry

Welcome, fellow traveler of the mathematical realm! Today, we are not just solving a problem; we are exploring the topography of a function.

We are given $f(x)={\int }_0^{x^2}{e}^{\sqrt{t}}(t-2)(t-3)dt$. At first glance, this integral might look like a mountain too steep to climb.

In JEE Advanced, we don't always need to conquer the mountain; sometimes, we just need to understand its peaks and valleys.

The Power of the Leibniz Rule

We want to know where this function increases, where it decreases, and where it hits its local maximums and minimums. To do this, we need the derivative, $f^{\prime }(x)$.

We invoke the Newton-Leibniz Rule. This rule is our secret weapon, stating that if we have a function defined by an integral with a variable upper limit $h(x)$, the derivative is the integrand evaluated at $h(x)$, multiplied by the derivative of $h(x)$.

Applying this to our function, where the upper limit is $x^2$, we get:

Simplifying this, we arrive at our derivative:

The Sign Analysis

Now, look at this expression. It is a product of three parts: $2x$, $e^x$, and the polynomial $(x^2-2)(x^2-3)$.

Since we are restricted to $x>0$, both $2x$ and $e^x$ are always positive. They are like the wind at our back—they don't change the direction of our journey.

The real action happens in the polynomial part. We find critical points where $f^{\prime }(x)=0$, which occurs when $x^2=2$ or $x^2=3$. Since $x>0$, our critical points are $x=\sqrt{2}$ and $x=\sqrt{3}$.

Let's map the behavior:

1. In the interval $(0,\sqrt{2})$: Here, $x^2<2$. Both $(x^2-2)$ and $(x^2-3)$ are negative. A negative times a negative is a positive, so $f^{\prime }(x)>0$, and our function is climbing.

2. In the interval $(\sqrt{2},\sqrt{3})$: Here, $2<x^2<3$. Now, $(x^2-2)$ is positive, but $(x^2-3)$ is negative. A positive times a negative is a negative, so our function is descending.

3. In the interval $(\sqrt{3},\infty )$: Here, $x^2>3$. Both brackets are positive. The function is climbing again.

The Final Synthesis

By the First Derivative Test, we have a local maximum at $x=\sqrt{2}$ because the function switches from increasing to decreasing.

We have a local minimum at $x=\sqrt{3}$ because it switches from decreasing to increasing.

Because $f^{\prime }(x)$ is continuous and differentiable, Rolle's Theorem guarantees that between our two critical points, there must be a point $c$ where the slope of the derivative is zero, meaning $f^{\prime \prime }(c)=0$.

The beauty of this problem lies not in the complexity of the integration, but in the elegance of the derivative's behavior. Keep this perspective, and you will find that even the most daunting calculus problems are just landscapes waiting to be mapped.