Question

Find the area of the region under the curve \(y=\) \(1 /\left(x^{2}+x\right)\) to the right of \(x=1\)

Short answer

The area is \( \ln(2) \).

Step-by-step solution

Understand the Problem

We need to find the area under the curve of the function \( y = \frac{1}{{x^2 + x}} \) from \( x = 1 \) to infinity. This can be considered an improper integral.

Set Up the Improper Integral

We set up the improper integral from \( x = 1 \) to infinity: \( \int_{1}^{ ext{∞}} \frac{1}{x^2 + x} \, dx \). Our goal is to evaluate this integral.

Factor and Simplify the Integrand

The integrand \( \frac{1}{x^2 + x} \) can be rewritten by factoring the denominator: \( \frac{1}{x(x+1)} \). This allows us to use partial fraction decomposition.

Partial Fraction Decomposition

Decompose \( \frac{1}{x(x+1)} \) into partial fractions: \( \frac{1}{x} - \frac{1}{x+1} \). Therefore, \( \int \frac{1}{x(x+1)} \, dx = \int \left( \frac{1}{x} - \frac{1}{x+1} \right) \, dx \).

Integrate Using Partial Fractions

Integrate term by term: \( \int \frac{1}{x} \, dx - \int \frac{1}{x+1} \, dx = \ln|x| - \ln|x+1| + C \). Simplifying, we get \( \ln\left|\frac{x}{x+1}\right| + C \).

Evaluate the Improper Integral

Evaluate the integral from 1 to \( b \) and then take the limit as \( b \) approaches infinity: \[ \lim_{b \to \infty} \left( \ln\left|\frac{b}{b+1}\right| - \ln\left|\frac{1}{1+1}\right| \right) \] Simplify to: \[ \lim_{b \to \infty} \left( \ln\left|\frac{b}{b+1}\right| - \ln\left(\frac{1}{2}\right) \right) \].

Calculate the Limit

As \( b \to \infty \), \( \frac{b}{b+1} \to 1 \), so \( \ln\left|\frac{b}{b+1}\right| \to \ln(1) = 0 \). Therefore, the expression simplifies to \[ 0 - \ln\left(\frac{1}{2}\right) = \ln(2) \].

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique that helps us break down complex rational expressions into simpler fractions, making them easier to integrate. In our problem, the function to integrate is \( \frac{1}{x^2 + x} \). To apply partial fraction decomposition, we first factor the denominator as \( x(x + 1) \).

Then, we express \( \frac{1}{x(x + 1)} \) as a sum of two simpler fractions:

• \( \frac{A}{x} \) and
• \( \frac{B}{x+1} \)

Next, we solve for constants \( A \) and \( B \) such that: \[ \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \] To find \( A \) and \( B \), we set the numerators equal to each other after clearing the denominators, which gives us the equation:
\[ 1 = A(x + 1) + Bx \]
Simplifying this, we find that \( A = 1 \) and \( B = -1 \), leading to the decomposition: \( \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x+1} \). This decomposition makes integration straightforward.

Area Under a Curve

Finding the area under a curve is a common application of integration, particularly when dealing with functions over specified intervals. In this context, we're focusing on the function \( y = \frac{1}{x^2 + x} \) and determining the area under it from \( x = 1 \) to infinity.

The notion of finding the "area under the curve" translates to calculating the definite integral of the function over that interval. Thus, we set up the integral: \[ \int_{1}^{\infty} \frac{1}{x(x+1)} \, dx \]

This is an improper integral because one of the limits of integration (infinity) is unbounded. Using techniques such as partial fraction decomposition helps to simplify the function into integrable parts. These parts, \( \frac{1}{x} \) and \( \frac{1}{x+1} \), are then individually integrated to help find the total area under the curve.

Limits at Infinity

Limits at infinity help us understand the behavior of a function as the input grows very large, typically towards infinity. When evaluating improper integrals, especially with limits that stretch to infinity, limits play a crucial role.

In our scenario, after integrating \( \frac{1}{x(x+1)} \) to \( \ln|x| - \ln|x+1| \), we evaluate the result from a finite point (\( x = 1 \)) to a varying endpoint \( b \), and then take the limit as \( b \to \infty \).

Thus, the expression becomes: \[ \lim_{b \to \infty} \left( \ln\left|\frac{b}{b+1}\right| - \ln\left(\frac{1}{2}\right) \right) \]

To compute this, we note that as \( b \) grows very large, \( \frac{b}{b+1} \) approaches 1, leading \( \ln\left(\frac{b}{b+1}\right) \) to approach \( \ln(1) = 0 \). Therefore, the remaining part \(- \ln\left(\frac{1}{2}\right) \) simplifies to \( \ln(2) \), which represents the area under the curve from 1 to infinity."}]}]} __[