Animated Solution for Mathematics - Definite Integration: Let the function $f:[0,2]\to \mathbb{R}$ be defined as $f(x)=\{\begin{array}{ll}{e}^{\min \{x^2,x-[x]\}},& x\in [0,1)\\ e^{[x-\ln x]},& x\in [1,2]\end{array}$, where $[t]$ denotes the greatest integer less than or equal to $t$. Then the value of the integral ${\int }_0^{2}xf(x)dx$ is

When you first encounter a problem like this, it is natural to feel a bit overwhelmed. You see a function defined in two different ways, a greatest integer symbol, a natural logarithm, and an exponential—it looks like a chaotic mess.

But in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

Decoding the First Interval $[0,1)$

Our function is defined as $f(x)=e^{\min \{x^2,x-[x]\}}$ for $x\in [0,1)$. The first thing to notice is the term $[x]$.

For any $x$ in the interval $[0,1)$, the greatest integer of $x$ is simply $0$. This is a huge relief! It means the exponent simplifies to $\min \{x^2,x-0\}$, which is just $\min \{x^2,x\}$.

Now, ask yourself: in the world of numbers between $0$ and $1$, which is smaller, $x$ or $x^2$? If you pick $0.5$, $x^2$ is $0.25$. If you pick $0.1$, $x^2$ is $0.01$.

The square is always smaller than the number itself. Therefore, $\min \{x^2,x\}=x^2$. Our function for the first interval is simply $f(x)=e^{x^2}$. That was not so scary, was it?

Decoding the Second Interval $[1,2]$

Now, let us look at the second part: $f(x)=e^{[x-\ln x]}$ for $x\in [1,2]$. This looks intimidating. Let us define $g(x)=x-\ln x$.

To understand its greatest integer, we need to know its range. We use the power of calculus: $g^{\prime }(x)=1-\frac{1}{x}$. Since $x\ge 1$, $\frac{1}{x}\le 1$, which means $g^{\prime }(x)\ge 0$. This tells us that $g(x)$ is a strictly increasing function.

The minimum value occurs at $x=1$, giving $g(1)=1-\ln (1)=1$. The maximum occurs at $x=2$, giving $g(2)=2-\ln (2)\approx 1.307$.

So, $g(x)$ lives between $1$ and $1.307$. The greatest integer of any number in this range is always $1$. Thus, $f(x)=e^1=e$ for the entire interval $[1,2]$.

The Integration

We have successfully simplified our function. The integral ${\int }_0^{2}xf(x)dx$ now splits into two manageable pieces:

Let us call these $I_1$ and $I_2$. For $I_1$, we use a simple substitution. Let $t=x^2$, then $dt=2xdx$, or $xdx=\frac{1}{2}dt$. When $x=0,t=0$; when $x=1,t=1$.

For $I_2$, since $e$ is a constant, we pull it out:

Adding them together, we get:

Look at that! We started with a complex piecewise function and ended with a clean, elegant result. This is the essence of JEE physics and math—breaking down the intimidating into the fundamental. Keep practicing, and you will find that every problem has a story waiting to be told.