Problem 69 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The area of the region enclosed by the curves \(y=x\), \(x=e, y=1 / x\) and the positive \(x\)-axis is (A) \(5 / 2\) square units (B) \(1 / 2\) square units (C) 1 square units (D) \(3 / 2\) square units

Step-by-Step Solution

Answer

The area of the enclosed region is \(\frac{3}{2}\) square units (Option D).

Step 1: Understand the Problem

We need to find the area of the region enclosed by the curves \(y = x\), \(y = \frac{1}{x}\), the line \(x = e\), and the positive \(x\)-axis. The area can be determined by integrating the difference between the functions over the appropriate interval.

Step 2: Find Points of Intersection

The curves \(y = x\) and \(y = \frac{1}{x}\) intersect when \(x = \frac{1}{x}\), leading to \(x^2 = 1\). Solving yields intersections at \(x = 1\) (a valid solution within the interval of interest).

Step 3: Set the Integration Limits

The region of interest is bounded between \(x = 1\) and \(x = e\). Hence, we will integrate from \(x = 1\) to \(x = e\).

Step 4: Set Up the Integral

Since \(y = \frac{1}{x}\) is above \(y = x\) in the region between \(x = 1\) and \(x = e\), the area under the curve can be expressed as the integral of \(\left(\frac{1}{x} - x\right)\) from \(x=1\) to \(x=e\).

Step 5: Calculate the Integral

Compute the integral \[\int_{1}^{e} \left(\frac{1}{x} - x\right) \, dx = \left[ \ln|x| - \frac{x^2}{2} \right]_{1}^{e}\]Evaluating this results in\[ (\ln e - \frac{e^2}{2}) - (\ln 1 - \frac{1}{2}) \]which simplifies to \[1 - \frac{e^2}{2} + \frac{1}{2}\].

Step 6: Simplify the Result

Simplifying the expression \[1 - \frac{e^2}{2} + \frac{1}{2} = \frac{3}{2}\].

Step 7: Choose the Correct Answer

After calculating, the area of the region enclosed by the curves is \(\frac{3}{2}\) square units, which corresponds to option (D).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that allows us to calculate the accumulated quantity, such as area under a curve, over a specific interval. Unlike indefinite integrals, which include an arbitrary constant, definite integrals provide a single answer by evaluating the antiderivative at the upper and lower bounds of the interval.
For instance, if you have a function \( f(x) \) and you want to find the area under this curve from \( x = a \) to \( x = b \), you set up a definite integral:

• Write the integral as \( \int_{a}^{b} f(x) \, dx \).
• Find the antiderivative \( F(x) \) of \( f(x) \).
• Evaluate \( F(x) \) at the upper limit \( b \), and subtract the evaluation at the lower limit \( a \): \( F(b) - F(a) \).

In the original exercise, we calculated the definite integral of \( \frac{1}{x} - x \) between \( x = 1 \) and \( x = e \) to find the area between two curves. This process exemplifies how definite integrals operate to give us precise, numerical values for areas.

Area Between Curves

The concept of the area between curves involves finding the space enclosed by two or more graphs. When two functions are defined over a certain interval, the area between them can be calculated by integrating the difference between the functions.
Here's how it works:

• Identify the top and bottom curves in the interval. In our exercise, the curve \( y = \frac{1}{x} \) is above \( y = x \) from \( x = 1 \) to \( x = e \).
• Set the integral of the difference between the top curve and the bottom curve: \( \int_{a}^{b} [\text{top function} - \text{bottom function}] \, dx \).
• Evaluate this definite integral to find the area between the curves.

This gives a quantitative measure of the region enclosed, considering the respective positions of the curves over the specific interval. Mastering this technique is crucial for solving various problems in calculus involving two or more graphs.

Intersection Points

Finding intersection points between curves is crucial when calculating areas or solving equations involving multiple graphs. These points indicate where the curves meet or overlap in a graph.
To find intersection points:

• Set the functions equal to each other: \( f(x) = g(x) \). In our example, this was \( x = \frac{1}{x} \).
• Solve the resulting equation for \( x \). For \( x^2 = 1 \), we found \( x = 1 \) as a valid solution.
• Ensure that the solutions lie within the interval of interest.

These points help to determine the limits of integration for areas between curves and indicate where one function transitions to being above or below the other. Identifying intersection points is critical for solving problems that involve evaluating definite integrals between boundaries given by the intersection of curves.