Problem 199 - A Complete Resource Book in Mathematics for JEE Main

Question

The area enclosed between the curve \(y=\log _{e}(x+e)\) and the coordinate axes is (A) 1 (B) 2 (C) 3 (D) 4

Step-by-Step Solution

Answer

The correct answer is (A) 1.

Step 1: Understand the Problem

We need to find the area enclosed between the curve \( y = \log_e(x+e) \) and the coordinate axes. This means we are interested in the area under the curve from where it intersects the x-axis to where it intersects the y-axis, for non-negative x and y.

Step 2: Identify Intersections

The curve \( y = \log_e(x+e) \) intersects the x-axis when \( y = 0 \). Solving \( \log_e(x+e) = 0 \) gives \( x+e = 1 \), so \( x = 1-e \). At y-axis, when \( x = 0 \), \( y = \log_e(e) = 1 \). Thus, we focus on the interval \( x = 1 - e \) to \( x = 0 \).

Step 3: Setup the Integral

We're finding the area under the curve from \( x = 1 - e \) to \( x = 0 \), so we set up the integral: \[ \int_{1-e}^{0} \log_e(x+e) \, dx \]

Step 4: Evaluate the Integral - Part 1 (Substitution)

Let \( u = x+e \), then \( du = dx \). Change of limits: When \( x = 1-e \), \( u = 1 \), and when \( x = 0 \), \( u = e \). The integral becomes \[ \int_1^e \log_e(u) \, du \]

Step 5: Evaluate the Integral - Part 2 (Integration by Parts)

Use integration by parts, with \( v = \log_e(u) \) and \( dw = du \). Then \( dv = \frac{1}{u} du \) and \( w = u \). This gives: \[ \int \log_e(u) \, du = u \log_e(u) - \int u \cdot \frac{1}{u} \, du = u \log_e(u) - u \]Evaluate this from 1 to e.

Step 6: Calculate Final Area

Now compute: - At \( u = e \), value is \( e \log_e(e) - e = e - e = 0 \).- At \( u = 1 \), value is \( 1 \cdot \log_e(1) - 1 = 0 - 1 = -1 \).The definite integral is \[ 0 - (-1) = 1 \].Thus, the area is 1.

Key Concepts

Area Under a Curve

When we talk about the 'area under a curve' in integral calculus, we are essentially referring to the integral of a function over a specified interval. This is a fundamental concept, as it provides us a way to calculate the total accumulation of a quantity, such as area, from a rate of change represented by a function.To find the area under the curve for the function given in the exercise (\( y = \log_e(x+e) \)), we identified the limits over which to integrate. The x-axis intersection at \( x = 1-e \) and the y-axis intersection at \( x = 0 \) defined our interval.By integrating from \( x = 1-e \) to \( x = 0 \), we calculated the area under the curve. The integration gives a precise measurement of this space, which is a crucial part of understanding how areas can be computed for complex curves.

Logarithmic Functions

Logarithmic functions are an essential topic in mathematics, particularly in integral calculus. A logarithmic function is the inverse of an exponential function. The function given in the exercise is \( y = \log_e(x+e) \), which means we are dealing with a natural logarithm.The base of the natural logarithm is 'e', an irrational and transcendental number approximately equal to 2.71828. Natural logarithms are common in calculus because they simplify differentiation and integration of exponential functions.In this exercise, understanding the behavior of \( \log_e(x+e) \) was crucial. It helped us identify where the curve intersects the axes and allowed us to set up necessary integral bounds. Logarithmic functions have unique properties, such as \( \log_e(1) = 0 \)and \( \log_e(e) = 1 \), which we used while evaluating the final integral.

Integration by Parts

Integration by parts is a method used in calculus to integrate products of functions. It's a technique based on the product rule of differentiation and is applicable when simpler methods don't suffice.The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]In our exercise, we implemented integration by parts to solve:\[ \int \log_e(u) \, du \]Here, we chose \( u = \log_e(u) \) and \( dv = du \), leading to \( v = u \)and \( du = \frac{1}{u} du \). By applying the integration by parts formula, we simplified to obtain the required integral result.Integration by parts is particularly powerful in turning complex logarithmic integrals into simpler expressions that can be evaluated directly.