Animated Solution for Mathematics - Definite Integration: Let $a$ and $b$ respectively be the points of local maximum and local minimum of the function $f(x)=2x^3-3x^2-12x$. If $A$ is the total area of the region bounded by $y=f(x)$, the $x$-axis and the lines $x=a$ and $x=b$, then $4A$ is equal to .

Unveiling the Area

Welcome, my dear student, to a beautiful exploration of calculus. Today, we are not just solving a problem; we are embarking on a journey to understand the geometry of a cubic function.

We are given the function $f(x)=2x^3-3x^2-12x$. At first glance, it is just a polynomial, but to a trained eye, it is a landscape of peaks and valleys waiting to be mapped.

Before we can calculate the area, we must define our boundaries. The problem asks for the area between the local maximum and the local minimum.

To find these, we look for the points where the slope of the tangent is zero. We compute the derivative:

Setting this to zero, we get $6(x^2-x-2)=0$, which factors beautifully into $6(x-2)(x+1)=0$. Thus, our critical points are $x=-1$ and $x=2$.

Using the second derivative test, $f^{\prime \prime }(x)=12x-6$, we find that at $x=-1$, $f^{\prime \prime }(-1)=-18<0$ (a local maximum), and at $x=2$, $f^{\prime \prime }(2)=18>0$ (a local minimum). These are our vertical boundaries.

Now, here is where many students stumble. We are tasked with finding the area bounded by the curve and the x-axis between $x=-1$ and $x=2$.

The naive approach is to simply compute ${\int }_{-1}^{2}f(x)dx$. But stop! If you do that, you are calculating the net signed area. If the curve dips below the x-axis, that area will be subtracted from the total.

We must check if the curve crosses the x-axis. Setting $f(x)=0$, we find $x(2x^2-3x-12)=0$. Lo and behold, $x=0$ is a root!

Since $0$ lies between $-1$ and $2$, our curve crosses the x-axis right in the middle of our integration interval. We must split our integral into two parts:

Let us perform the integration. The antiderivative of $f(x)$ is:

For the first region, $A_1$:

For the second region, $A_2$:

The physical area is the absolute value, $|-24|=24$.

We have our two pieces: $A_1=4.5$ and $A_2=24$. The total area $A$ is their sum:

But wait! The question asks for $4A$. We must not stop now. We multiply our result by $4$:

There it is—the elegance of the final answer, 114. You have navigated the derivative, identified the critical points, avoided the trap of the x-intercept, and performed the integration with precision. This is the essence of JEE mathematics: patience, visualization, and rigorous execution.