Animated Solution for Mathematics - Definite Integration: Let for $x\in \mathbb{R}$, $f(x)=\frac{x+|x|}{2}$ and $g(x)=\{\begin{array}{ll}x,& x<0\\ x^2,& x\ge 0\end{array}$. Then area bounded by the curve $y=(f\circ g)(x)$ and the lines $y=0,2y-x=15$ is equal to ______.

Welcome, fellow traveler of the mathematical realm. Today, we are not just solving a problem; we are embarking on a journey to visualize the hidden architecture of functions.

We have been given a composite function $y=(f\circ g)(x)$ and asked to find the area it bounds with a line and the $x$-axis. It sounds daunting, but let us break it down, step by step, with the precision of an architect and the curiosity of an explorer.

Decoding the Gatekeeper

First, let us look at $f(x)=\frac{x+|x|}{2}$. This is not just an algebraic expression; it is a fundamental operator.

Think of it as a 'gatekeeper.' When $x>0$, $|x|=x$, so $f(x)=\frac{x+x}{2}=x$. When $x<0$, $|x|=-x$, so $f(x)=\frac{x-x}{2}=0$.

In the world of neural networks, this is known as the ReLU function. It is a simple, elegant switch: if you are positive, you pass through; if you are negative, you are silenced. This realization is our first victory.

The Composite Dance

Now, we introduce $g(x)$. We are told $g(x)=x$ for $x<0$ and $g(x)=x^2$ for $x\ge 0$. We need to find $f(g(x))$.

If $x<0$, then $g(x)=x$, which is negative. Our gatekeeper $f$ sees this negative input and returns $0$.

If $x\ge 0$, then $g(x)=x^2$, which is always non-negative. Our gatekeeper $f$ sees this positive input and lets it pass through unchanged. Thus, our composite function is:

We have successfully simplified a complex composition into a clean, piecewise function.

Visualizing the Boundaries

Now, let us look at the boundaries. We have the curve $y=f(g(x))$, the $x$-axis ($y=0$), and the line $2y-x=15$.

Rearranging the line gives us $y=\frac{x+15}{2}$. To find the bounded region, we need to know where these lines meet.

Setting $x^2=\frac{x+15}{2}$ for $x\ge 0$ leads us to the quadratic equation:

Factoring this, we get $(2x+5)(x-3)=0$. Since we are in the positive domain, $x=3$ is our intersection point. At $x=3$, $y=9$.

The line also hits the $x$-axis at $x=-15$. We now have a clear picture: a triangle-like region bounded by a line on top and a combination of the $x$-axis and a parabola on the bottom.

The Elegant Solution

Here is where we use our intuition. We could integrate the piecewise function, but why do that when we can use geometry?

The area under the line $y=\frac{x+15}{2}$ from $x=-15$ to $x=3$ forms a large right-angled triangle. The base is $18$ units (from $-15$ to $3$) and the height is $9$ units (at $x=3$).

The area of this triangle is:

However, this triangle includes the area under the parabola from $x=0$ to $x=3$, which is not part of our bounded region. So, we calculate the area under the parabola:

Subtracting this from our triangle area, we get $81-9=72$.

And there we have it. By breaking down the functions, visualizing the geometry, and using the subtraction of areas, we have arrived at the answer of 72.

This problem was never about brute-force calculation; it was about understanding the behavior of functions and the elegance of geometric interpretation. Keep this mindset, and no problem will ever be too complex for you.