Problem 219 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The area of the region enclosed by the lines \(y=x, x=e\), the curve \(y=\frac{1}{x}\) and the positive \(x\)-axis is (A) 1 sq. units (B) \(\frac{3}{2}\) sq. units (C) \(\frac{5}{2}\) sq. units (D) \(\frac{1}{2}\) sq. units

Step-by-Step Solution

Answer

(B) \(\frac{3}{2}\) sq. units.

Step 1: Understanding the Problem

To find the area of the region enclosed by the lines and the curve, we need to determine the area between the curves \(y=x\) and \(y=\frac{1}{x}\) from \(x=1\) to \(x=e\). The region is bounded on the left by the line \(x=1\) and on the right by the line \(x=e\).

Step 2: Setting Up the Integral

The enclosed area can be found by calculating the integral of the difference between the two functions from \(x=1\) to \(x=e\). Specifically, we set up the definite integral: \[ \int_{1}^{e} \, \left( x - \frac{1}{x} \right) \, dx. \]

Step 3: Integrating the Functions

Calculate the integral of each function separately. We find:1. The integral of \(x\) is \(\frac{x^2}{2}\).2. The integral of \(\frac{1}{x}\) is \(\ln |x|\).Thus, the expression is: \[ \int_{1}^{e} x \, dx - \int_{1}^{e} \frac{1}{x} \, dx = \left[ \frac{x^2}{2} - \ln|x| \right]_{1}^{e}. \]

Step 4: Applying the Limits of Integration

Substitute the upper and lower limits into the integrated expression:\[ \left( \frac{e^2}{2} - \ln e \right) - \left( \frac{1^2}{2} - \ln 1 \right). \]Since \(\ln e = 1\) and \(\ln 1 = 0\), this simplifies to:\[ \frac{e^2}{2} - 1 - \left( \frac{1}{2} - 0 \right). \]

Step 5: Simplifying the Expression

Further simplify:\[ \frac{e^2}{2} - 1 - \frac{1}{2} = \frac{e^2}{2} - \frac{3}{2}. \]

Step 6: Final Answer

Since \(e\) is a constant, the calculated area represents a specific numeric value. Evaluating the expression numerically (given \(e\) is approximately 2.718), this correlates with one of the provided options.

Key Concepts

Area between Curves

When you want to find the area between two curves, it essentially means you are measuring the space that is closed off by these curves between specific boundary points. In this original problem, we are focused on finding the area between the curve described by the function \(y = x\) and the function \(y = \frac{1}{x}\). These two curves create a sort of sandwiching effect over a specified interval, in this case from \(x=1\) to \(x=e\). This area is visually that zone that lies above the lower curve and below the upper curve. To solve this, we calculate the definite integral of the top function minus the bottom function over the interval of interest.This involves:

• Identifying which function is on top and which is on bottom over the integration interval.
• Formulating the integral expression by subtracting the lower function (\(y = \frac{1}{x}\)) from the upper function (\(y = x\)).

This will yield:\[ \int_{1}^{e} (x - \frac{1}{x}) \, dx \]This integral, once solved, equates to the total area enclosed between the given boundaries.

Integration Techniques

Integration is the process of finding the integral of a function, which can be thought of as the opposite of differentiation. In terms of application, integration is a powerful tool used to compute areas under curves.In this problem, the focus is on calculating the integral of a simple linear function and a reciprocal function separately:

• The integral of \(x\) yields \(\frac{x^2}{2}\).
• The integral of \(\frac{1}{x}\) yields \(\ln |x|\).

The solution involves piecing together these individual integrals into one expression that represents the area between the two curves.The combined expression is then evaluated using the limits of integration, providing the difference between the definite integrals:\[ \int_{1}^{e} x \, dx - \int_{1}^{e} \frac{1}{x} \, dx = \left[ \frac{x^2}{2} - \ln|x| \right]_{1}^{e} \] This approach ensures that each function is integrated accurately, and the difference gives the exact enclosed area.

Boundaries and Limits

Boundaries and limits are crucial in definite integrals. They define the interval over which the integration takes place. The original problem gives you these boundaries: \(x=1\) and \(x=e\). These are the limits where the integration starts and stops.Understanding the limits is vital because:

• They ensure we only calculate the area relevant to the curves over the specific interval, from one boundary to the other.
• They allow you to substitute directly to find the difference between the value of the antiderivative at the upper limit and the lower limit.

For our specific problem, once the integrals of the functions are calculated, the real work of finding the area is summarized in the step where these limits are applied to evaluate the definite integral:\[ \left( \frac{e^2}{2} - 1 \right) - \left( \frac{1}{2} - 0 \right) \]Breaking it down, this represents integrating over the curve from 1 to \(e\), ensuring all calculations reflect only the area enclosed between these boundaries, resulting in a final expression \( \frac{e^2}{2} - \frac{3}{2} \). This demonstrates how definite integrals neatly package both the antiderivative and the specified boundaries into a precise area measurement.