Animated Solution for Mathematics - Definite Integration: The area of the bounded region enclosed by the curve $y=3-\left|x-\frac{1}{2}\right|-|x+1|$ and the x-axis is

A Journey into the Piecewise

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a jagged, intimidating mountain range of absolute values.

The function $y=3-|x-\frac{1}{2}|-|x+1|$ might seem complex, but I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for underlying symmetry and elegance.

Our goal today is not just to find the area, but to see the hidden geometry that the algebra is trying to whisper to us.

The Surgery of Critical Points

Before we can draw or calculate anything, we must perform a little 'surgery' on the function. Absolute value functions are piecewise by nature; they change their behavior at the points where the expression inside the bars hits zero.

These are our critical points. Look at the terms: $|x-\frac{1}{2}|$ and $|x+1|$.

Setting $x-\frac{1}{2}=0$ gives us $x=\frac{1}{2}$. Setting $x+1=0$ gives us $x=-1$.

These two values, $x=-1$ and $x=\frac{1}{2}$, are the 'hinges' of our graph. They divide the entire real number line into three distinct acts of a play. We must analyze the function in each act separately to understand its true form.

The Three Acts

In this region, both $(x-\frac{1}{2})$ and $(x+1)$ are negative. When we remove the absolute value bars, we must negate the expressions to keep the result positive.

Our function becomes:

Simplifying this, we get $y=3+x-\frac{1}{2}+x+1$, which simplifies beautifully to $y=2x+\frac{7}{2}$.

To find where it hits the x-axis, we set $y=0$, leading us to $2x=-\frac{7}{2}$, or $x=-\frac{7}{4}$. We have found our first vertex: $(-\frac{7}{4},0)$.

Now, we enter the middle ground. Here, $(x-\frac{1}{2})$ is still negative, but $(x+1)$ has become positive.

The function transforms into:

Watch closely as we expand this: $y=3+x-\frac{1}{2}-x-1$. The $x$ terms cancel out!

We are left with $y=\frac{3}{2}$. This is the 'Plateau'—a perfectly horizontal line segment. It connects the points $(-1,\frac{3}{2})$ and $(\frac{1}{2},\frac{3}{2})$.

Finally, for $x>\frac{1}{2}$, both expressions are positive. The function becomes:

Simplifying this yields $y=\frac{5}{2}-2x$. This is a line with a negative slope, descending back toward the x-axis.

Setting $y=0$, we find $2x=\frac{5}{2}$, so $x=\frac{5}{4}$. Our final vertex is $(\frac{5}{4},0)$.

The Geometric Revelation

Now, step back and look at what we have built. We have a shape with vertices at $A(-\frac{7}{4},0)$, $B(-1,\frac{3}{2})$, $C(\frac{1}{2},\frac{3}{2})$, and $D(\frac{5}{4},0)$.

Because the segment $BC$ is horizontal ($y=\frac{3}{2}$) and the segment $AD$ lies on the x-axis ($y=0$), these two lines are parallel!

A quadrilateral with a pair of parallel sides is a trapezoid. We don't need calculus; we need the geometry of our ancestors.

The Final Calculation

The area of a trapezoid is given by the formula:

1. Base 1 ($b_1$): The distance from $A(-\frac{7}{4},0)$ to $D(\frac{5}{4},0)$ is $\frac{5}{4}-(-\frac{7}{4})=\frac{12}{4}=3$. 2. Base 2 ($b_2$): The distance from $B(-1,\frac{3}{2})$ to $C(\frac{1}{2},\frac{3}{2})$ is $\frac{1}{2}-(-1)=\frac{3}{2}$. 3. Height ($h$): The vertical distance between $y=0$ and $y=\frac{3}{2}$ is simply $\frac{3}{2}$.

Plugging these into our formula:

And there it is. The area is $\frac{27}{8}$ square units.

You didn't just solve a problem; you mapped a landscape. Keep this intuition—that algebra is just a tool to describe the shapes of the universe—and you will conquer any problem the JEE throws your way.