Animated Solution for Mathematics - Definite Integration: Let $f:[-3,1]\to \mathbb{R}$ be given as $f(x)=\{\begin{array}{ll}\min \{(x+6),x^2\},& -3\le x\le 0\\ \max \{\sqrt{x},x^2\},& 0\le x\le 1\end{array}$. If the area bounded by $y=f(x)$ and $x$-axis is $A$, then the value of $6A$ is equal to ____.

Imagine you are standing on the coordinate plane, looking at a function that refuses to be defined by a single rule. This is the beauty of piecewise functions—they are like a story with different chapters, each with its own character.

Our function $f(x)$ is defined as $\min \{(x+6),x^2\}$ for $x\in [-3,0]$ and $\max \{\sqrt{x},x^2\}$ for $x\in [0,1]$. To find the area $A$ bounded by this curve and the $x$-axis, we must first master the art of the 'switch'.

In the first interval, $[-3,0]$, we are looking for the minimum of a line, $y=x+6$, and a parabola, $y=x^2$. To understand where the function switches from the line to the parabola, we must find where they meet.

We set $x+6=x^2$, which rearranges into the quadratic equation:

Factoring this gives us $(x-3)(x+2)=0$. We have two potential intersection points: $x=3$ and $x=-2$.

Since our interval is restricted to $[-3,0]$, we discard $x=3$ and focus solely on $x=-2$. Before this point, from $x=-3$ to $x=-2$, the line $x+6$ is the lower value. After this point, from $x=-2$ to $x=0$, the parabola $x^2$ dips below the line.

Moving to the second interval, $[0,1]$, we are tasked with finding the maximum of $\sqrt{x}$ and $x^2$. Here, we rely on our intuition about numbers between $0$ and $1$.

When you square a fraction like $1/2$, you get $1/4$, which is smaller. When you take the square root of $1/4$, you get $1/2$, which is larger.

Thus, for the entire interval $[0,1]$, $\sqrt{x}$ is consistently greater than or equal to $x^2$. Our function simplifies beautifully to $f(x)=\sqrt{x}$ in this region.

Now that we have our clear, three-part definition: $f(x)=x+6$ for $x\in [-3,-2]$, $f(x)=x^2$ for $x\in [-2,0]$, and $f(x)=\sqrt{x}$ for $x\in [0,1]$, we can proceed to the integration. The total area $A$ is the sum of three integrals:

Calculating $I_1={\int }_{-3}^{-2}(x+6)dx$, we find the result is $7/2$. For $I_2={\int }_{-2}^0{x}^{2}dx$, the result is $8/3$. Finally, for $I_3={\int }_0^1\sqrt{x}dx$, we get $2/3$.

Adding these together:

The journey is almost complete, but remember the golden rule of JEE: always answer the specific question asked. We found $A=41/6$, but the question asks for $6A$.

Multiplying our result by $6$:

It is a moment of pure mathematical satisfaction when the denominators cancel out, leaving us with a clean, elegant integer. You have successfully navigated the piecewise landscape, and the final answer is 41.