Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_{e^{-1}}^{e^2}|\frac{{\log }_{e}x}{x}|dx$ is

A Journey Through the Integral

Welcome, future IITians! Today, we are not just solving a math problem; we are embarking on a journey to understand the behavior of functions. We are tackling the definite integral $I={\int }_{e^{-1}}^{e^2}\left|\frac{\ln x}{x}\right|dx$.

This problem is a classic in the JEE Advanced repertoire because it tests your ability to handle the 'modulus'—a function that demands respect and precision. Let us break this down step by step.

The Detective Work

Whenever you see an absolute value sign inside an integral, your first instinct should be to find where the expression inside changes its sign. Think of the modulus function $|f(x)|$ as a mirror. If the function is positive, the mirror does nothing; if the function is negative, the mirror flips it above the x-axis.

Our function is $f(x)=\frac{\ln x}{x}$. We are operating on the interval $[e^{-1},e^2]$. Since the denominator $x$ is always positive in this range, the sign of the entire fraction depends solely on the numerator, $\ln x$.

We know that $\ln x=0$ when $x=e^0=1$. This point, $x=1$, is our critical transition point. To the left of $1$, the natural log is negative; to the right, it is positive. This is the moment where we must split our path.

The Piecewise Strategy

With our critical point identified, we can define our modulus function piecewise. For $x\in [e^{-1},1)$, $\ln x<0$, so $\left|\frac{\ln x}{x}\right|=-\frac{\ln x}{x}$. This negative sign is crucial—it mathematically reflects the curve from below the x-axis to above it.

For $x\in [1,e^2]$, $\ln x\ge 0$, so the absolute value simply outputs $\frac{\ln x}{x}$. Now, we use the additive property of definite integrals to split our main integral into two distinct parts:

Let us call these $I_1$ and $I_2$. $I_1$ represents the area of the 'flipped' region, and $I_2$ represents the area of the naturally positive region.

The Elegant Substitution

Now, we enter the heart of the calculation. Notice the structure of our integrand: we have $\ln x$ and its derivative, $\frac{1}{x}$, sitting right there. This is a perfect setup for substitution.

Let $t=\ln x$. Differentiating both sides with respect to $x$, we get $\frac{dt}{dx}=\frac{1}{x}$, which implies $dt=\frac{1}{x}dx$.

We must change our limits of integration to match our new variable $t$: - When $x=e^{-1}$, $t=\ln (e^{-1})=-1$. - When $x=1$, $t=\ln (1)=0$. - When $x=e^2$, $t=\ln (e^2)=2$.

Our integrals are now transformed into something much cleaner:

The Final Calculation

Let us compute $I_1$. The integral of $-t$ is $-\frac{t^2}{2}$. Applying the limits from $-1$ to $0$:

Now for $I_2$. The integral of $t$ is $\frac{t^2}{2}$. Applying the limits from $0$ to $2$:

Finally, we combine our results:

What a beautiful result! We have navigated the modulus, performed a clean substitution, and arrived at the solution. Remember, in JEE Advanced, it is not just about getting the answer; it is about the elegance of the process.