If ∫_(1/3)³ |logₑ x| dx = (m/n) logₑ (n²/e), where m and n are coprime natural numbers, then m² + n²

Q: Why do we have to split the integral at $x=1$? Can't we just integrate it directly?

Great question! The modulus function $|\ln x|$ is defined piecewise: it is $-\ln x$ when $\ln x < 0$ (which happens when $0 < x < 1$) and $\ln x$ when $\ln x \ge 0$ (which happens when $x \ge 1$). Because the definition of the function changes at $x=1$, we cannot use a single antiderivative for the entire interval $[1/3, 3]$. Splitting the integral is the only way to correctly account for the reflection of the negative part.

Q: I'm getting confused with the negative signs when evaluating the first integral. How do I avoid mistakes?

It is a classic trap! When you integrate $-\ln x$, keep the negative sign outside the entire evaluation bracket: $-[x \ln x - x]_{1/3}^{1}$. Evaluate the definite integral $[x \ln x - x]_{1/3}^{1}$ first, and only after you have that result, distribute the negative sign. Taking it step-by-step like this prevents sign errors.

Q: How do I transform my final result into the form $\frac{m}{n} \ln (\frac{n^2}{e})$?

Once you have your numerical result, say $A \ln 3 - B$, look at the target form. You need a logarithm of a quotient. Use the property $k \ln x = \ln (x^k)$ to bring the coefficient inside the logarithm, and remember that $1 = \ln e$. This will allow you to use the quotient rule $\ln a - \ln b = \ln (a/b)$ to reach the required structure.

Question

If $\int_{1/3}^{3} |\log_e x| dx = \frac{m}{n} \log_e (\frac{n^2}{e})$, where $m$ and $n$ are coprime natural numbers, then $m^2 + n^2 - 5$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to evaluate the integral of the absolute value of the natural logarithm:
$$I = \int_{1/3}^{3} |\ln x| dx$$

The function $y = \ln x$ is a smooth curve that crosses the $x$-axis at $x = 1$. For $x < 1$, the values are negative, and for $x > 1$, the values are positive.

When we introduce the modulus sign, $|\ln x|$, we reflect the negative portion of the curve across the $x$-axis. This creates a sharp, $V$-like corner at $x = 1$, which serves as the critical point for our integration.

### The Art of Splitting

Because the function changes its behavior at $x = 1$, we must partition the integral into two distinct regions. We integrate the negative of the function where it is below the axis, and the function itself where it is above.

The integral becomes the sum of two parts:
$$I = \int_{1/3}^{1} (-\ln x) dx + \int_{1}^{3} (\ln x) dx$$

By negating the first part, we effectively "flip" it to positive, satisfying the definition of the modulus function.

### The Integration Engine

We utilize the standard integral result derived from integration by parts:
$$\int \ln x dx = x \ln x - x + C$$

First, we calculate the integral for the interval $[1/3, 1]$:
$$I_1 = -[x \ln x - x]_{1/3}^{1}$$

Substituting the limits:
$$I_1 = -[(1 \ln 1 - 1) - (\frac{1}{3} \ln(1/3) - \frac{1}{3})]$$
$$I_1 = -[-1 - (-\frac{1}{3} \ln 3 - \frac{1}{3})] = -[-1 + \frac{1}{3} \ln 3 + \frac{1}{3}] = \frac{2}{3} - \frac{1}{3} \ln 3$$

Next, we calculate the integral for the interval $[1, 3]$:
$$I_2 = [x \ln x - x]_{1}^{3}$$
$$I_2 = (3 \ln 3 - 3) - (1 \ln 1 - 1) = 3 \ln 3 - 3 - (-1) = 3 \ln 3 - 2$$

### The Final Transformation

Adding the two components together, we obtain:
$$I = (\frac{2}{3} - \frac{1}{3} \ln 3) + (3 \ln 3 - 2)$$
$$I = \frac{8}{3} \ln 3 - \frac{4}{3}$$

To match the target form $\frac{m}{n} \ln (\frac{n^2}{e})$, we factor out $\frac{4}{3}$:
$$I = \frac{4}{3} (2 \ln 3 - 1) = \frac{4}{3} (\ln 9 - \ln e) = \frac{4}{3} \ln \left( \frac{9}{e} \right)$$

Comparing this to the form $\frac{m}{n} \ln (\frac{n^2}{e})$, we identify $m = 4$ and $n = 3$. These values are coprime, satisfying the problem constraints.

Finally, we calculate the requested value:
**$m^2 + n^2 - 5 = 4^2 + 3^2 - 5 = 16 + 9 - 5 = 20$**

Animated Solution for Mathematics - Definite Integration: If $\int_{1/3}^{3} ∣ lo g_{e} x ∣ d x = \frac{m}{n} lo g_{e} (\frac{n ^{2}}{e})$ , where $m$ and $n$ are coprime natural numbers, then $m^{2} + n^{2} - 5$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Visualizing the Modulus Function

Splitting the Integral at $x = 1$

The Integration Tool

Evaluating the First Part: Upper Limit

Evaluating the First Part: Lower Limit

Result of the First Integral

Evaluating the Second Part: Upper Limit

Evaluating the Second Part: Lower Limit

Combining Both Results

Matching the Required Form

Final Logarithmic Simplification

The Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Art of Splitting

The Integration Engine

The Final Transformation

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