Problem 58 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\int_{0}^{1} \frac{d x}{2 e^{x}-1}=p \log (q e-1)-r\), then (A) \(p=1, q=1, r=-1\) (B) \(p=1, q=2, r=1\) (C) \(p=1, q=2, r=-1\) (D) None of these

Step-by-Step Solution

Answer

(C) \(p=1, q=2, r=-1\)

Step 1: Analyze the Integral

We need to evaluate the integral \( I = \int_{0}^{1} \frac{d x}{2 e^{x}-1} \). The function inside the integral is \( \frac{1}{2 e^{x}-1} \), which requires a substitution to simplify it further.

Step 2: Performing Substitution

Let's substitute \( u = 2 e^{x} - 1 \). Then \( du = 2 e^{x} \, dx \), which implies that \( dx = \frac{du}{2 e^{x}} = \frac{du}{u+1} \) since \( e^x = \frac{u+1}{2} \). When \( x = 0 \), \( u = 1 \), and when \( x = 1 \), \( u = 2e - 1 \).

Step 3: Rewrite the Integral

With the substitution, the integral becomes \( I = \int_{1}^{2e-1} \frac{du}{u(u+1)} \). We can simplify this using partial fraction decomposition.

Step 4: Partial Fraction Decomposition

Express \( \frac{1}{u(u+1)} \) as partial fractions: \( \frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} \). Solving for \( A \) and \( B \), we find \( A = 1 \) and \( B = -1 \). Thus, \( \frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1} \).

Step 5: Integrate the Expression

Now integrate \( \int_{1}^{2e-1} \left( \frac{1}{u} - \frac{1}{u+1} \right) du = \left[ \log|u| - \log|u+1| \right]_{1}^{2e-1} \).

Step 6: Evaluate the Definite Integral

Calculate the definite integral: \( \left[ \log(2e-1) - \log(2e) + \log(1) - \log(2) \right] = \log(1 - \frac{1}{2e}) \).

Step 7: Simplify the Expression

Now, \( \log(1 - \frac{1}{2e}) = - \log(\frac{2e}{2e-1}) \). The integral can be expressed as \( -\log(\frac{2e}{2e-1}) = \log(\frac{1}{2e}) \), thus the expression matches the form \( p \log(qe - 1) - r = \log(\frac{1}{2e}) \).

Step 8: Match with Given Expression

The form matches with \( p = 1, q = 2, r = -1 \), which corresponds to option (C).

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus that helps simplify complex integrals by changing variables. This method works by transforming the original integral into a form that's easier to solve, utilizing a new variable. In this particular exercise, the substitution was used to deal with the integral \( \int \frac{d x}{2 e^{x}-1} \). The complex function \( \, \frac{1}{2 e^{x}-1} \, \) inside the integral is simplified by substituting \( u = 2 e^{x} - 1 \). This substitution transforms \( dx \) using the derivative \( du = 2 e^{x} \, dx \), leading to \( dx = \frac{du}{u+1} \).
When working with the substitution method, it is important to adjust the limits of integration according to the new variable. Here, the limits change from \( x = 0 \) and \( x = 1 \) to \( u = 1 \) and \( u = 2e - 1 \) respectively. This transformation is key to evaluating the integral in a more manageable form, allowing further techniques like partial fraction decomposition.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions.
These simpler fractions can individually be integrated more easily. In this example, after using substitution, the integral takes the form \( \int_{1}^{2e-1} \frac{du}{u(u+1)} \).
To simplify, it's essential to express \( \frac{1}{u(u+1)} \) as a sum of fractions:\[ \frac{1}{u} - \frac{1}{u+1} \].
Finding the constants \( A \) and \( B \) involves setting an equation \( \frac{A}{u} + \frac{B}{u+1} \), which when solved here gives \( A = 1 \) and \( B = -1 \).

• Partial fractions allow the integration to proceed separately on each term, simplifying the original integral into pieces that are easier to handle.

This decomposition reveals an obtainable path to solve the integral, allowing students to proceed with integration term by term.

Calculus

Calculus provides the tools needed to compute the integral and solve real-world problems involving rate of change and accumulation of quantities. \( \int_{0}^{1} \frac{d x}{2 e^{x}-1} = \log(\frac{1}{2e}) \) is an example of using definite integrals to find an exact result between specific limits.
The fundamental theorem of calculus links the concept of differentiation and integration, allowing the solution of integrals to provide a function's accumulated total change.

• This highlights calculus's ability to solve problems involving continuous change, like the continuously growing exponential function \( e^x \), which appears frequently in real-world scenarios such as population growth and compound interest.
• In this exercise, integrating \( \frac{1}{u} \) and \( -\frac{1}{u+1} \) results in the equation \( \log|u| - \log|u+1| \), which is evaluated with the specified bounds to derive the results.

Understanding these operations ensures comprehensive solutions, bridging steps to create a cohesive process for various calculus applications.